New Batches at TathaGat Delhi & Noida!               Directions to CP centre
Permutation and Combination
by Total Gadha - Sunday, 17 August 2008, 11:13 AM
  cat 2009 cat 2010 permutation and combination Dagny has been telling me smugly that it's now my turn to write an article. So I decided to write one as soon as possible and pass the buck back to her. We both know that it takes a lot of pain to research for an article and write the text so we both try to avoid it for as long as possible. On the other hand, since we both enjoy writing and like to publish a well-finished chapter, we both undertake the task eagerly once the ball is in our court. Writing these chapters have made me realize how true is the maxim "if you want to learn something well, teach it." I can cite a lot of topics which I learnt while teaching in the class or answering students' questions. These years of teaching have made so many topics seem pretty childish to me which were nightmares to me before. But knowing a topic well and teaching it are two different ball games. Many a times, I have found myself struggling in the class to make a student understand something which I have found obvious to understand! The fault is not on the student's part. Three or four years ago, I would have been stuck on the same point! Every time I encounter a situation like this, my only measure is to stop at that point, retrace my steps with the students and slowly unravel the difficulty he's facing. It works most of the times. But sometimes, the student is in so much awe with the topic that his mind gets frozen. I have seen this happening many a times. Topics such as time, speed and distance, Permutation and Combination,  geometry etc. inspire so much fear among the students that sometimes simple principles, which they would have otherwise understood do not strike them as simple. I get incessant queries such as "Sir, time speed distance chhod sakte hain?," "Sir, permutation combination na karein to chalega?" from my students all the time. And the sad part is, that the level of CAT in these topics is very simple. They are doable. In the present chapter, I am trying to present one of these dreaded chapters as I understand it. I hope my students understand it too.

P&C1

P&C2
P&C3
P&C4
P&C5
P&C6
P&C7

I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.

 

If you think this article was useful, help others by sharing it with your friends!


Bookmark and Share
You might also like:
Probability
Maxima, Minima and Inequalities

Re: Permutation and Combination
by vamsi krishna - Sunday, 17 August 2008, 12:23 PM
 

Sir,

Tatz a GEM OF AN ARTICLE

thnkz,

VaMsI

Re: Permutation and Combination
by Amit A - Sunday, 17 August 2008, 05:22 PM
  Now what do i say other than "THANKS A LOT TG !!! "
u rock ....


Amit
Re: Permutation and Combination
by Amit Kumar - Sunday, 17 August 2008, 06:14 PM
 

Very Very useful article..

Thanks

Re: Permutation and Combination
by Leonidas S - Sunday, 17 August 2008, 06:20 PM
 

Hi TG,

A fabulous article.I think there is a mistake in the solution of Prob 3 of the 4 friends problem.

It says "The total number of ways in which 4 friends can stay together is

104   " I think it should be 10000 as you found out earlier.smile

Re: Permutation and Combination
by Amit Kumar - Sunday, 17 August 2008, 09:40 PM
 

Hi TG,

Answer to 'Good Boy Bad Boy' movie question:

Since these two couples want to sit together at either end. Now, we'll tie them together. Hence, total number of students will be 7 out of which 6 students can sit in 6x5x4x3x2x1 = 720 ways. and the couple can sit at either ends (i.e 2 ways) and the couples can sit in 2 ways among themselves and finally, in each couple, boy and girl can sit in 2 ways(BG and GB)(Total = 2x2 ways). Hence, Number of ways all can sit will be:  

720x2x2x2x2 = 11520 (Answer)

Re: Permutation and Combination
by Manika Tandon - Sunday, 17 August 2008, 10:41 PM
 

hey TG...

awesome piece of writing dude... rather thanx dagny for giving him a push to write it... :D

both of u all rock..

cheers,

smile

Re: Permutation and Combination
by AsHwIn Drmz - Sunday, 17 August 2008, 11:02 PM
  hi....Tg....

G8 article...thk u so much for a wonderful article,actually i was worried as u were not writing articles for cat 08 students....but nw m so glad and confident dat my all worries will vanish.Tg can u plz write an article on probability too....smile
Re: Permutation and Combination
by Ashwin A - Monday, 18 August 2008, 07:51 AM
  TG Sir!
Simply Fantabulous!! Just wanted to know do u have a Book having all topics of Quant...I have the book on Number theory which in itself, is very good.!!!
Re: Permutation and Combination
by Dagny Taggart - Monday, 18 August 2008, 09:17 AM
  Hey Manika,

TG played smart this time. I asked him to write an article three days back ;he wrote, edited and uploaded it within two days. Now the ball is back in my court. angry
Re: Permutation and Combination
by Diwakar Bhaskar - Monday, 18 August 2008, 09:47 AM
  Thnks a lot TG...........
i dont have word to say anything more ...................
Re: Permutation and Combination
by vamsi krishna - Monday, 18 August 2008, 11:13 AM
 

Hey Amit,

Pl Recheck

Shudn't Ans be 720 x 2 x 2 x 2

1..six students can be arranged in 6! = 720

2..The 2 couples can sit at either of the ends in 2 ways

3..and then each couple on either side can exchange their seats..in 2 ways so two couples can do tat in 2 X 2 ways

regards,

VaMsI

Re: Permutation and Combination
by Krushang Shah - Monday, 18 August 2008, 11:36 AM
  THANKS A  LOT TG!!!

its a wonderful article.....
Re: Permutation and Combination
by amit kumar - Monday, 18 August 2008, 01:44 PM
 

Thanks TG

Awesome Article.......

Re: Permutation and Combination
by nikhil khandelwal - Monday, 18 August 2008, 04:58 PM
 

Thanks a lot TG

Doing a great job

Re: Permutation and Combination
by rajdeep choudhuri - Monday, 18 August 2008, 05:46 PM
 

Hi TG,

The article is simply mind blowing!!!Plzzzzzzzz post an exercise on Permutation and Combination problems.

Re: Permutation and Combination
by bryan cole - Tuesday, 19 August 2008, 12:59 AM
  Is the answer to a+2b+3c.....total number of elements is 36?
Re: Permutation and Combination
by Total Gadha - Tuesday, 19 August 2008, 03:37 AM
  Hi Rajdeep,

About the exercise, have you seen the article completely? evil


Total Gadha
Re: Permutation and Combination
by rajdeep choudhuri - Tuesday, 19 August 2008, 04:22 PM
 

Yessmile.But i feel it will be lot more helpful if you can give us an exercise like the one on Number System problems,I mean  plenty more problems to solve.Also where can i get the answers to these problems???

Anywys TG rocks!!!!cool

Re: Permutation and Combination
by Riyaz Iqbal - Wednesday, 20 August 2008, 11:03 PM
  Hi TG,

    Excellent article,as usual.I've a question about the 'counting functions' problem.The no.of functions from a 5-element set to 7-element set should be less than 7^5,right? 7^5 simply represents the number of ordered pairs (x,y) from first set to the second.Several of them may correspond to a single function.eg: (1,1) (2,2),...(5,5) are five distinct ordered pairs but a single function viz y=x. Am I missing something?
Re: Permutation and Combination
by ands ands - Thursday, 21 August 2008, 07:22 AM
 

ash, hi I've just started with the Num system but am finding the area vast,enomously huge and the topics not concentrated from a particular source.From your post I see that there is some consolidated resource for the Num system you talkin about or is the the ebook by TG ppl.Can you post on the name here.It would be sure help if I am to find the book and the contents satisfying.Thanks.....!!!!

Re: Permutation and Combination
by Total Gadha - Thursday, 21 August 2008, 01:18 PM
  Hi Riyaz,

There are 75 arrangements, and every arrangement, mapping 5 elements of set 1 to 7 elements of set 2 would be counted as a single function only.

Total Gadha
Re: Permutation and Combination
by Shalini B - Thursday, 21 August 2008, 02:45 PM
  Hi TG,

I have registered for your Copycat exams by paying through credit card but I am facing problems while trying to write the exam online. I have mailed you long back about this (to the copycat email ID and also to admin email ID). But I have not received any answer for that. And I could not find any contact number too on this site where I could call up and report my problem. Please respond.

 Content from my previous mail dated July 30th 2008 to you below:

My problem is that when I tried taking copycat 2 the site was unresponsive and did not navigate to the next page at all. I tried taking the exam but when I started with it and entered my options for the first page and clicked on the Next button, the site became very slow and did not show the next page at all. I tried several times but it was of no use. I checked the browser compatibility test and it said my system is fine for taking the test and my internet connection is good and all other websites are working fine. I am mighty disappointed. Please look into this problem ASAP
Re: Permutation and Combination
by manoj 1123 - Thursday, 21 August 2008, 04:06 PM
  Hi TG,
Though I feel not ok to write here, I am also facing the same problem, not able to browse to next page. may be if the page remains open for some time the connection to ur server breaks.
Please send the pdf for the COPYCATs to my mail beforehand.
So that I can take print and appear the test..
Plz find a solution for us.

Thanks.
Re: Permutation and Combination
by Leonidas S - Friday, 22 August 2008, 10:33 PM
  Hi TG,

That was an excellent article

Can you please post some articles on functions,logarithms and probability


Re: Permutation and Combination
by rakesh ojha - Sunday, 24 August 2008, 08:02 PM
 

hi dear

First of all ..thnx a lot for such a nice collections of questions which has helped to build the fundamentals of this topic...

 

i just have one doubt in one of the examples ..above where we are getting number of 7 didgit numbers ...in binary.....that part is ok...

but when the number is converted to decimal base....there will be other conditions aprt from last 3 digits as zeros........say a binary no when conveted to decimal and becomes 24...that is divisble by 8....

may be i m wrong somewhere but plz tell me why only you have taken the condition of last 3 didgits as 0 for dibisiblity by 8...why not other conditions....

 

thnx

rakesh

Re: Permutation and Combination
by Total Gadha - Sunday, 24 August 2008, 10:12 PM
  Hi Rakesh,

Convert 24 to binary please?

Total Gadha
Re: Permutation and Combination
by amit kumar - Monday, 25 August 2008, 01:31 AM
  HI ALL,
question.>there are 10 students out of which three are boys and seven are girls,in how many different ways can the students be paired such that no pair consists of two boys?

Unable to solve this.Please help wid proper explanation
Re: Permutation and Combination
by IRONMAN A - Monday, 25 August 2008, 11:57 AM
  Thanks a million for the beautiful article TG.

I have a small doubt. Plz do explain me.

"There are 6 tasks & 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned either to person 3 or person 4. Every person must be assigned one task. In how many ways can the assignment be done(CAT 2006)?"

What's the wrong in my procedure....

First person: As task 1 & task 2 cannot be assigned any of the other 4 tasks can be assigned to him - 4 ways
Second Person: As task 1 & task 2 cannot be assigned, and one already assigend to Person 1,  any of the other 3 tasks can be assigned to him - 3 ways
Third Person & Fourth person: one of these get Task 2. The other gets any of the other 3. (one of tasks 1,3,4,5,6 (leaving the tasks assigned to Persons 1 & 2) - 3 ways
Fourth & Fifth Persons: other 2 tasks can be assigned in 2 ways.

total = 4*3*3*2 = 72

Plz do answer me.. thanks in advance

Re: Permutation and Combination
by arpi sin - Monday, 25 August 2008, 12:20 PM
  TG/Dagny
Wonderful work!!!
I was going through the number system quiz,and have a doubt in a question .Where can I post my query ?

Regards
Arpita
Re: Permutation and Combination
by Surendran Chandravathanan - Monday, 25 August 2008, 09:43 PM
 

Hi Amit,

Here's my try...

1st Approach

1 boy can pair up wid 7 girls in 7 ways...So for 3 boys, total no. of ways is 21..And 1 girl can pair up wid 6 other girls in n(n-1)/2, ie, (7x6)/2 = 21 ways..(or) Just 7C2 = 21 will cede u the answer.

So the final ans is 21 + 21 = 42 ways..

[[OR]]

2nd Approach

Should this problem be solved like below....???

To select 2 ppl from 10, we can do in 10C2 = 45 ways..And arranging them in pairs can be done in 45x2 = 90 ways..(or) juz we have to arrange 2 places out of 10 which can be done in 10P2 = 90 ways..

And then, 3 boys can be paired up in 6 ways..

So is the final ans 90 - 6 = 84 ways ??????

I am perplexed..

Which ans is correct, TG ????? Appreciate ur comments.........

Best Rgds,

Suren

Re: Permutation and Combination
by Surendran Chandravathanan - Monday, 25 August 2008, 09:58 PM
 

TGGGGGG,

Ur art of solving & simplifying the problems leaves me in awe but it invites more confusions.....

I wud say dat the below problem can be solved easily but u said it is not so..

In how many ways can four persons be seated out of 5 boys and 3 girls on four different seats?

Now see, totally thr are 8 ppl where they 've to be seated in 4 seats..

So won't the normal permutation formula do ???????

ie, 8P4 = 1680 ways....

Plz elucidate the fact for breaking the problem & considering 4 cases in ur approach....

Thanks in advance,

Best Rgds,

Suren

Re: Permutation and Combination
by Black i - Monday, 25 August 2008, 11:38 PM
  Hi TG,
It is regarding the '7 digit binary no div. by 8 when converted to decimal no' question.
How can one decide that the last 4 bits , if are equal to zero then the no is divisible by 8 when converted to decimal ?
Suppose the same question asks about octal no instead of binary, how should one decide about it ?
Re: Permutation and Combination
by amit kumar - Tuesday, 26 August 2008, 10:21 PM
  Hi
Answer is 630.give another try.
Re: Permutation and Combination
by Harish Bansal - Thursday, 28 August 2008, 12:51 PM
  Hi TG,
Gr8 article....
Plz post the solutions of the problems that you have mentioned in ur artical...
We want to match our answers and approach.
And plz can u post something on probability prior to CAT 2008.

Eagrly waiting for a reply.........


Re: Permutation and Combination
by Tuhin Banerjee - Tuesday, 2 September 2008, 05:14 PM
 

Hi TG, it's realy a mind blowing article on permutation and combination.

I have question on the same topic and have two different approaches , kindly tell me which one correct :

Question: Ten students have been shortlisted to form two teams of six students each, such that there are exactly three common between the two teams, In how many ways can the teams be formed.

 Ans:

1st approach:

Three common stuednts we can take out of 10 students  = 10 Cways    ,

       We need to build each group having 6 students , for the 1stGR. rest of the 3 students we can take out of 7 students =7C3 ways

So , 1stGR complete, for the 2ndGR we can take rest of the 3 students out of the 4 students in = 4Cways.

  Total = 10 C* 7C3  * 4C3   = 16800

 

2nd Approach :

From 10 students , 3 students commonbetween the two teams can be selected in = 10C3 ways.

since  7 students are still left and for each team we need 6 students , Now select the student

will  find a place in neither team and it can be done in = 7C1 ways.

Now remaining 6 students have to divided into two groups of 3 each and it can be done in

 = 6C3 / 2

Now two teams can be formed = 10C3 * 7C1 * 6C3 / 2 = 8400

please tell me where I am missing the concepts smile

 

 

Re: Permutation and Combination
by rudra samal - Tuesday, 2 September 2008, 10:01 PM
 

Hi TG,

In the example where u asked:

How many arrangements r dere of six 0's, five 1's and four 2's, where

(i)first 0 precedes first 1?

shudnt the ans be 343980 and not 349380?.... and there's an easier way as well. It goes like this:

total arrangements=(15!/(4!*5!*6!))

so ans will be =total arrangements*(p/q)

where q=all possibl arrngmnts of the 1s and 0s

           p=arrngmnts where 0 occupies 1st position

so,

q=11!/(6!*5!)

p=10!/(5!*5!)

ans=343980

cool

Re: Permutation and Combination
by Total Gadha - Wednesday, 3 September 2008, 10:49 AM
  Hi Rudra,

Corrected the typing error. And good way to solve the question. smile

Total Gadha
Re: Permutation and Combination
by Total Gadha - Wednesday, 3 September 2008, 10:51 AM
  Hi Tuhin,

The second solution is correct. First has repetition. Please check the question of "three states with three students from each state...."

Total Gadha
Re: Permutation and Combination
by bhageerath n - Wednesday, 3 September 2008, 10:30 PM
  hi
i'm here for the first time.. this was pretty helpful.. how do i cross check if i got the right answers?? will the solutions be available for the questions??
Re: Permutation and Combination
by bhageerath n - Wednesday, 3 September 2008, 11:04 PM
  Hi TG,
for the question -
In how many ways can four persons be seated out of five boys and three girls on 4 different seats?

should we take those cases as mentioned by you??
why cant we do it like this?

there are 8 people in all.. select four of them in 8c4 ways (70 ways). Now you can arrange these four in 4! ways(24 ways). So total ways is 70*24 = 1680.

is there anything wrong with this approach??
Re: Permutation and Combination
by Syd S - Thursday, 4 September 2008, 01:58 PM
  kudos for another brilliantly structured and edited article,,,!!!!cool

was just wondering guys bout dis prob,,,
"in how many diff ways can the faces of a cube be painted in 6 diff colors?"

shudnt the no of ways be 6 X 5 x 3!,,
as the first face can be painted in 6 ways,,
rather than only 1 way,,

 jus wanted 2 knw,,y is it 5 X 3! rather than 6 X 5 x 3!

thnx
Re: Permutation and Combination
by brijesh gogia - Saturday, 6 September 2008, 07:12 PM
  Thank You for this very useful article.
Re: Permutation and Combination
by ganesh p - Sunday, 7 September 2008, 08:41 AM
  That U Very much sirsmile..........as expected my search for the best article on permutation and combination did end here..
Re: Permutation and Combination
by kladad asda - Sunday, 7 September 2008, 08:44 PM
  just being curious. are u vamsee krishna from CSE IITKGP, class 2006??
If you are kindly reply back, it would be nice to know one more familiar face at TG
Re: Permutation and Combination
by Manish Kashyap - Tuesday, 9 September 2008, 07:56 PM
 

Hi TG,

    My Solutions are -

1)a

2)c

3)b

4)

5)c

6)

7)c

8)c

9)a

10)b

11)b

12)c

13)b

14)b

15)

16)

17)b

19)d

20)

21)c

22)16725
23)24
24)840

25)120

26) 680
27)26!/2
28)45

29)21

30)35
31)5
32)7c2
33)33600
I was not able to solve some questions for which I had not put any answers. Can you please verify my answers and tell me the solutions of which are wrong or which I was not able to solve.

Thanks and Regards,

Manish

 

 

Re: Permutation and Combination
by Ankit Arora - Wednesday, 10 September 2008, 05:27 PM
 

Hi TG,

 

I haven't solved all the questions till now...just want to confirm the answers of first 5 ques..

1 (a)

2(b)

3(d)

4. I m getting (26^3)*220 as my answer

5(c)

 

Re: Permutation and Combination
by Manish Kashyap - Thursday, 11 September 2008, 03:09 PM
 

Hi TG Sir,

Please check the solutions pleazzzzzzzzzzzzzzz. I am waiting for solutions.

Thanks and Regards,

Manish

Re: Permutation and Combination
by vamsi krishna - Friday, 12 September 2008, 04:25 PM
 

Frns

 

Check/discuss answers for above problems @

http://totalgadha.com/mod/forum/discuss.php?d=3683

 

VaMsI

Re: Permutation and Combination
by Saurabh Shekhar - Friday, 26 September 2008, 09:59 AM
  Hi TG,

Could you please post the complete solution to the Exercise.

We are all eagerly waiting for it.

Saurabh
Re: Permutation and Combination
by Venkkatesan R - Saturday, 27 September 2008, 04:44 AM
  Hi suren,

I think the question involves just the selection. Hence the arrangement is not required. Hence 10c2-3c2=45-3=42 is the answer.
Re: Permutation and Combination
by arvind jenamani - Saturday, 4 October 2008, 09:19 PM
 

Hi TG sir,

The link is fabulous. In combination part there is a question " In how many ways 5 boys , 3 girls can be sitted in 4 seats?"

There we don't need to segregate into parts like you mentioned.

8C4 ways we can select 4 people(without considering boy,girl combination)

4! ways to arrange them. so, 8C4*4!=8P4=1680

which match also your answer.

 

 

Re: Permutation and Combination
by Amit Trivedi - Saturday, 11 October 2008, 10:31 PM
 

Hi TG sir,

The lesson is really great. i have a doubt for one of the examples.

the examplein which 5 boys and 5girls go to watch the cinema spiderman3.

you gave two arrangement for the wish of the boys "every boy wants to sit with a girl" and no two girls sit together

i think other arrangement of the form are possible

G B B G B G B G B G

G B G B B G B G B G

G B G B G B B G B G

G B G B G B G B B G

In the above arrangements every boy is sitting with a girl and no two girls are sitting together

clarify my doubt, let me know where i am making mistake in reasoning.

Warm Regards,

Amit

Re: Permutation and Combination
by Amit Trivedi - Saturday, 11 October 2008, 11:57 PM
 

if the couples are siting at the ends

the arangements are 2 X 2 X 2 x 6!

Amit

Re: Permutation and Combination
by meena k - Saturday, 18 October 2008, 04:24 AM
 

hi,

for the states question can't we solve like this from 3states 3 students eachso selecting 5 frm 9= 9c5

then subtracting 6c5 (case where all 5 students are selected frm two states)

ie:9c5 - 6c5 .pls tell me where the problm is

 

Re: Permutation and Combination
by noel sandesh - Monday, 3 November 2008, 03:40 PM
 

Hi TG,

This is in regard to an example problem (No. of 7 digit binary numbers that are divisible by 8 when converted to base 10). i was wondering as to y the 3rd digit from the last cannot be 1 or 0. You've mentioned that the last 3 digits should be 0. But if the last 3 digits are 100 then its base 10 equivalent will be 400 which is divisible by 8.... I'm really bad at P n C but i can't figure out the logical flaw in my thought...  Can u help me out on this?

Re: Permutation and Combination
by rajesh kambampati - Thursday, 6 November 2008, 12:36 AM
  Ya i agree with you sandesh, i too got this doubt.And i think we are correct.
Re: Permutation and Combination
by Total Gadha - Friday, 7 November 2008, 12:49 AM
  (100)2 = 4 (in decimal)
Re: Permutation and Combination
by Complete Gadha - Friday, 7 November 2008, 08:03 PM
  Hi TG Sir,

This is the best material i have ever come across for Permutation and Combination.

I have been going through the material in the forum. all of them r so amazing that i am actually gaining the confidence to crack the CAT though i have not been doing so good in the mock cat. Thanks a lot..!!
Hats off To U...!!!
Re: Permutation and Combination
by Bheemesh K - Wednesday, 12 November 2008, 01:40 AM
  Hi TG,

I had a doubt in the first problem (i.e 10 speakers problem) solved in the permutation section.

I solved it in this manner. you tie the PM, MP and MLA into a single person and then calculate the ways to arrange them.
wudn`t it be 8! then that we have other 7 speakers and 1 group of PM, MP and MLA to arrange?

I find your way also logical but am not able to understand difference between my approach and your approach. kindly can you explain where I am missing the logic?

Re: Permutation and Combination
by sumit jamwal - Wednesday, 4 February 2009, 08:51 AM
 

Hi TG,

I have some problem in understanding the concept of partition problem..

of 5 balls as stated be you.

 let's take  * * * * * these as 5 similar balls

now dividing them into 3 grps..

i can do it by placing 2 sticks ..for that i have total of

|*|*|*|*|*|    --> | denotes choices

6 choices..

so selecting 2 spaces out of 6 can be done in

6C2 = 15 ways..

tel me wats goin wrong in this approach..

 

Re: Permutation and Combination
by sumit jamwal - Thursday, 12 February 2009, 09:41 PM
  hi TG,
the problem i had above , i'd done a mistake
actually thr wud be 30 ways ...

for 1st stick - 6 ways
for 2nd stick - 5
so total - 6*5=30.

if you cud clarify this thing it wud be nicesmile

Regards,
sumit
Re: Permutation and Combination
by Total Gadha - Thursday, 12 February 2009, 11:09 PM
  Hi Sumit,

In your solution, you can't place sticks like this for example
||* * * * *

This would denote a 0 0 5 case which you cannot cover. You can only cover 0 5 0 case by placing the sticks in this manner: |* * * * *|

The way you are placing the sticks you can never place two sticks side by side because you are taking it to be one space only.

Total Gadha
Re: Permutation and Combination
by sumit jamwal - Friday, 13 February 2009, 12:16 AM
  hi sir,
sir that's fine ..bt if iam nt considering this case ..then y is my answer comin to be arnd 30 ..whr as u stated it's 21.
i think thr might be somekind of repeatition...

and as u told
| | * * * * * nt included so,


now suppose i place 1st stick
then arrangement wud be like..
| * * * * *

now to include ur case for 2nd stick thr are 6 cases as i can place
stick between stars and beside 1st stick

in this case total cases come out to be 36  still more than 21

a bit confused as how to get correct result by using this stick method.

regards,
sumit
Re: Permutation and Combination
by Total Gadha - Friday, 13 February 2009, 09:32 AM
  Hi Sumit,

The repetition is because you are taking partitions to be different whereas there have to be taken as similar. For example, take two simple cases:

* P1 * * * P2 *  and * P2 * * * P1 * where P1 and P2 denote the partitions. Both these case would yield the same result, i.e. 1 3 1 but you are counting them as different.

If I do this by your method I will do it like this:
For the first partition I have 6 places.
* * | * * *

For the second partition I have 7 places. Therefore total number of ways = 6 × 7 = 42. As each case is being repeated once, the total number of cases = 42/2 = 21.

Total Gadha
Re: Permutation and Combination
by sumit jamwal - Friday, 13 February 2009, 05:07 PM
 

hi TG,

thankscool,that was wat i thinking....

gr8 work ..

Regards ,

sumit

Re: Permutation and Combination
by Abhinav Kumar - Friday, 22 May 2009, 01:38 PM
 

Thats really a beautiful one.

Made such a difficult task so simple with the examples of such a variety.

Thanks a Lot

                            Abnv

 

 

Re: Permutation and Combination
by rony roy - Friday, 22 May 2009, 02:33 PM
  hi TG,

what was the answer to the question that goes.....

how many arrangements of six 0s five 1s and four 2s are there in which

ii)the first 0 precedes the first 1,(and) precedes the first 2.

my answer comes out to be a massive : 504504

reply soon...
Re: Permutation and Combination
by iim freak - Tuesday, 26 May 2009, 12:29 PM
 

hi tg.. 1st of all thanks a ton for this simply awesome post .. i ws nt getting any gud material on P&C n Probability n was worried ... thanks again..

.. regarding the Problem :: there are 10 speakers who r supposed 2 address. One constraint is PM shud address before MP and MP before MLA .

My soln:: Let PM = P ; MP = Q; MLA = R

the order shud be PQR .. so i tuk PQR as 1 block and so total people 2 b arranged becomes (7+1) = 8 ... and now arranging them we get

Ans = 8! .

Please tell where am i making the mistake ?

 

 

Re: Permutation and Combination
by Priyesh Tungare - Wednesday, 27 May 2009, 02:59 PM
  Hi TG,

Thanks for the article.smile

I have one doubt.

In the question, How many 3-digit numbers are even and with no repeated digits, in the second case for number not ending in 0, you have multiplied 8*8*4.

Can you explain why we are multiplying 4 here?

This may be simple question for others, bt i believe in saying.. when in doubt ask....

Thanks,
Priyesh Tungare

Re: Permutation and Combination
by Total Gadha - Thursday, 28 May 2009, 02:06 PM
  Hi Randeep,

'Before' might not always mean 'just before,' as you have assumed by taking them together. There can be other persons between P, Q and R.

Total Gadha
Re: Permutation and Combination
by Total Gadha - Thursday, 28 May 2009, 02:08 PM
  Hi Priyesh,

There are four 4 even digits after removing zero- 2, 4, 6 and 8. The number would end in one of these.

Total Gadha
Re: Permutation and Combination
by iim freak - Thursday, 28 May 2009, 10:21 PM
 

oh yeah ,, gosh i shud hav thot dat.. thanks a lot for clearing the doubt ...

TG sir it will b a gr8 help if u cud post some gud text, like dis, for Probablity also .. dats another area, which i find tough..

Re: Permutation and Combination
by iim freak - Friday, 29 May 2009, 10:09 AM
  hi tg.. from where can i find the answers 2 the  above exercise.. its v imp 2 chk whether i hav understud ur concepts pr nt.
Re: Permutation and Combination
by anirban bhar - Sunday, 31 May 2009, 12:34 PM
  hi tg,
im a bit confsed wid the solution to the paint cube problem............

the firstface wich we choose cn b painted in 6 ways,.........y havent u considered that....plzzz enlighten
Re: Permutation and Combination
by Avishek Chakraborty - Thursday, 11 June 2009, 07:39 PM
 

Hey guys I have a question related to this topic, could u  solve this? Here it goes...

Q. 4 boyz n 4 girls are arranged in a row so that no 2 girls are together.Wat is the probability that any 2 boyz are together?

waiting for ur response...

Re: Permutation and Combination
by iim freak - Thursday, 11 June 2009, 08:27 PM
  dude .. this is PnC thread n u asking Probability ..jst kidding ..  anyways i dnt know probability so cant help @ this moment .. TG sir will be posting a text on Prob soon... waiting for that...ATB mate
Re: Permutation and Combination
by ku klux klan - Thursday, 18 June 2009, 07:18 PM
  there are 4 friends and 10 hotel rooms
q)in how many ways can at least two friends stay together

my answer to this is as follows,which is leading me to the wrong answer,will be very grateful if someone can point out the fallacy in this

1 case)-2 friends stay together
c(4,2)*10*9*8

2nd case)three friends stay together

c(4,3)*10*9

3rd case)all four stay together

10

adding up all these cases i am getting 4690,which is wrong....please tell me the mistake in this
Re: Permutation and Combination
by rohan kaushal - Saturday, 20 June 2009, 09:03 AM
  hi,

Q.   how many three digit numbers are there which are even and have no repeated digits ????



can't it be this way???

_ _ _
the 3rd place 5 choices ,(i.e; 0,2,4,6,8) so no of ways are 5
1st place have 8 choices (excluding 0 and the 3rd place's number)
& 2nd place have again eight choices ....
and that comes out to be  8 * 8 * 5 = 320 ..

could you plz help me .......
thanx a lot!!!!
Re: Permutation and Combination
by amit jain - Sunday, 21 June 2009, 12:37 PM
 

@Rohan,

When you will use 0 for the unit's place then for the hundreds place you will have 9 choices(1....9) and 1oth place 8 choices.

I hope now you know where you are getting it wrong smile

Re: Permutation and Combination
by srinivasan ravi - Sunday, 12 July 2009, 10:51 AM
  hi sir,
can u pls explain me the problem just before cicular combinations..i cant understand it clearly..pls help..thanks..
Re: Permutation and Combination
by smruty panigrahi - Monday, 13 July 2009, 11:10 AM
 
Hii,

This is a rather simple looking problem but has been pestering me and am unable to solve. Pl chech it out and help me guys.

You have to find out the number of ways to reach from A to B.

Extremely sorry 4 the poor quality of fig. I ain't that good with computers.

Plz give a detailed solution.

Thanx in advance. smile
Re: Permutation and Combination
by aryan raj - Monday, 13 July 2009, 03:03 PM
 

hi

sir

this file is so good for cat exam so i want solution of the all prob

so plz sand me the solution on my email

Re: Permutation and Combination
by Ankit Megotia - Tuesday, 28 July 2009, 11:13 AM
  hey smruty,
by any chance do you know the answer to your question?
its coming out to be 72.
if its correct i will tell you the procedure.
Re: Permutation and Combination
by Jaideep Das - Tuesday, 28 July 2009, 10:44 PM
 

Can somebody provide me the answers for (ii) and (iii) parts of the question for 6 0's, 5 1's ans 4 2's?

I got the answers as 15840 for (ii) and 378738 for (iii)

Are these answers correct? Please reply

Re: Permutation and Combination
by Gowtham Muthukkumaran Thirunavukkarasu - Monday, 3 August 2009, 11:20 PM
  You r right.
The solution given here, is based on a assumption that the three speek in order, which is not the case. Your solution is perfect i guess.
Re: Permutation and Combination to Bheemesh K
by Gowtham Muthukkumaran Thirunavukkarasu - Monday, 3 August 2009, 11:22 PM
  You r right.
The solution given here, is based on a assumption that the three speek in order, which is not the case. Your solution is perfect i guess.
Re: Permutation and Combination TO smruty panigrahi
by Gowtham Muthukkumaran Thirunavukkarasu - Tuesday, 4 August 2009, 12:31 AM
 
From the figure you have given i have worked out and got the answer to be 30. Juz check if you already know the answer and if it is wrong please let me know. I have attached the file here.


Moreover I have assumed that one can travel through a path only once. For example you cannot go through ED and again comeback to E.
Re: Permutation and Combination
by Avishek Chakraborty - Wednesday, 5 August 2009, 03:31 PM
 

Firstly i must say the article is awesome!

Please pardon me if my question seems childish!

There's one question related to dice, in how many ways sum of 8 can be obtained by rolling 2 dice if the they are distinguishable...

now my question is as we are rightly considering (3,5) and (5,3) as different cases, why are we not taking two cases for (4,4)? Though the denominations are same but the dice are distinguishable.

Re: Permutation and Combination
by cat champ - Saturday, 8 August 2009, 04:11 PM
  i dont think it will be taken abhishek.. as(4,4) is making the dice indistinguishable..

btw.. very good work tg sir..the best article till date i have found on p&c.. and i will be posting the solns soon..
Re: Permutation and Combination
by animesh chandan - Saturday, 15 August 2009, 10:57 PM
  hi TG
i hav problem
Q:there are 7 men and 5 women we have to select 4 men and 3 women but if mrs A will go then mrs B will not go.in how many ways v can select them?

my approach is 7 men will be selected in 7C4 ways
and there are 3 cases
case 1: we first select mrs A them automatically mrs b will not be selected then we can select remaining women in 3C2 ways,
that is 7C4*3C2=105.
case 2: we select mrs B then automatically mrs A will not be selected then ve can select remaining women in 3C2 ways,
that is 7C4*3C2=105,

case 3:if both of them are not selected then wewill have 3C3 ways to select women
that is 7C4*3C3=35
that means total no of ways =105+105+35=245

plZZZ tel me is it the right approach?????
Re: Permutation and Combination
by Anoop Maithani - Thursday, 20 August 2009, 10:30 AM
  Hi TG..

Amazing article ..waiting eagerly for an article on Probability..
Re: Permutation and Combination
by BellTheCAT ... - Thursday, 20 August 2009, 07:18 PM
 

Hi TG,

A really helpful article,for it helped me revise P&C. thanks a lot.

Re: Permutation and Combination
by deep sin - Saturday, 5 September 2009, 01:59 PM
  @ ku klux klan

You are forgetting to take one case into calculation

The case is : - 2 stay in same hotel and remaining 2 also stays in same hotel.
Re: Permutation and Combination
by Pallav Jain - Thursday, 17 September 2009, 07:03 PM
 

Hi All,

Can you plz tell me the solution ?

Using one or more of the digits 1, 2, 3, 4, 5, 6 and 7, how many 7-digit numbers can be formed which are divisible by 7?

Options :-

A) 75 + 74 + 73 + 72 + 7           C) 4 × 76

B) 4 × 75                                       D) 76                  E) 6 × 76

Thanks

Pallav Jain

Re: Permutation and Combination
by priya ramraj - Friday, 18 September 2009, 04:00 PM
  Hi TG,
I have a doubt in the article that u have posted. To get a sum of 8 with distinguishable dices, wont the occurrence of(4,4) be different because it is distinguishable??? shudnt we consider it as 2 cases ..Pls clarify...
Re: Permutation and Combination
by ROHIT K - Friday, 18 September 2009, 05:05 PM
  Hi priya,

Jus to giv it a try..

To get a sum of 8 with distinguishable dices, wont the occurrence of(4,4) be different because it is distinguishable??? shudnt we consider it as 2 cases ..


(2,1) and (1,2) are different in distinguishable dice(which is not the case in same kind of dice).

Say, you have a red and a blue die

Case: (1,2) and (2,1)

1.Red->1 Blue->2

2.Red->2 Blue->1

You can distinguish the two not only by their color but also by the difference in their numerical value.

Case 2: (4,4)

1.Blue->4 Red->4
2.Red->4 Blue->4

Aren't they both same??

If yes, then we need not consider it two cases.

In short, since the numerical value is same, we cannot distinguish the case though the two dice are distinguishable.

Hope this helps.smile

Rohit

P.S if you still didn't understand...TG to hai..!!
Re: Permutation and Combination
by priya ramraj - Friday, 18 September 2009, 05:31 PM
  Hi Rohit,
going by ur example,Wat i understand from the article is that
color is the distinguishing factor and not the number... If we remove the color aspect from the argument only then the distinguishing property vanishes... hope i am not wrong..anyways.. Thanks for the approach
Re: Permutation and Combination
by ROHIT K - Saturday, 19 September 2009, 02:17 PM
  Hi priya,

Color is the distinguishing factor and not the number...

If we remove the color aspect from the argument only then the distinguishing property vanishes...


I do not completely agree with your conclusion..

The dice had two different colors viz. red and blue..(I did not remove them)

But, still you could not distinguish <4,4> from each other..!!(how could you if they are the same ones?)

The reason was not color but same numerical value..
If you had changed the number to <1,2> then you would have distinguished it from <2,1> occurrence.

if you can check my explanation again after reading this hope it would be more clear..

1. If colors of dice are same(i.e indistinguishable dice) then occurrence of <1,2> or <2,1> or for that matter any pair of values and vice versa cannot be distinguished..(very obvious) (indistinguishable because of color)

2. If colors of dice are different then

a. <1,2> can be distinguished from occurrence of <2,1>(distinguishable because of color of the dice)

b. <4,4> or for that matter any same value on both dice cannot be distinguished from vice versa value(which is actually the same occurrence as the previous..) (indistinguishable because of same value on the dice)

Hence, the conclusion can be drawn that the distinguishing factor in the dice of different colors is  the color aspect, only when numbers appearing on the dice are different. If the number becomes same(indistinguishable) then you cannot distinguish the occurrence.

Hope this helps..

Rohit

Re: Permutation and Combination
by Suresh S - Monday, 21 September 2009, 11:11 AM
  Hi sir ..
its really superb.... thanks for ur article.....
Re: Permutation and Combination
by ROHIT K - Tuesday, 22 September 2009, 12:36 PM
  Hi TG..!!

Thanks for such a nice article.

I had one question regarding:

In how many ways can 4 persons be seated out of 5 boys and 3 girls in 4 different seats?

While explaining you said it is not the simple formula based P&C problem.

Though, I think that it could be solved in very simple terms without breaking down the problem into cases(the way you already suggested as the concept).

Two aspects to this problem:

1. Selecting 4 persons from 8 persons(5 boys+3 girls) which can be done in 8C4 ways

2. Arranging the 4 persons on 4 different seats for each selection which can be done in 4! ways

So total number of ways = 8C4 * 4! = 70*24 = 1680   (same as 8P4)

Please let me know if this is not the right way to approach the problem.
I need your insight into this.

Thanks and Regards

Rohit
Re: Permutation and Combination
by Rohan S - Tuesday, 29 September 2009, 11:49 AM
  TG STANDS FOR TRUE GUIDE........
Re: Permutation and Combination
by sharmishtha Gupta - Friday, 9 October 2009, 01:13 PM
 

Hi,

 

superb article.

i have a doubt though,

in the ques where u hv 10 college students , 5 boys and 5 girls and u have to make sure that no two girls sit together,

can we first make the boys sit in 5! ways

  _B_B_B_B_B_

now, for the girls to sit , we have to choose 5 of the 6 places available so that no two girls sit  together, this can be done in 6 ways and the girls can then be seated in 6* 5! ways.

Therefore, the no of ways in which they can sit according to the given condition then comes out to be

6*5!*5!.

 

Please correct me if i am wrong.

Thanx

Re: Permutation and Combination
by AsHwIn Drmz - Saturday, 10 October 2009, 01:32 AM
 

hi Sharmista,

You are missing out some hig here. Say 1st we seat all the boys..in 5! ways.

Then we have 6 places where 5 girls have to be seated.Now, 1st girl can be seated in any of the 6 available places,2nd girl can be seated in the remaining 5 avaialble places ,3rd girl can be seated in remaining 4 available places....etc

ie 6 x 5 x 4 x 3 x 2 .

so total arrangements boils down to : 2 * 5! *(6x5x4x3x2)

Hope its clear now.

Ashwin

Re: Permutation and Combination
by sharmishtha Gupta - Saturday, 10 October 2009, 10:03 PM
  Thanx ashwin.

U might think me to be a dud but kindly explain how u got the

2 in 2*5!*(6*5*4*3*2) .
Re: Permutation and Combination
by AsHwIn Drmz - Saturday, 10 October 2009, 10:59 PM
 

hey..

dont think like dat..We are all here to learn from eachother no one is perfect.So never underestimate urself...Coming to the question, see there will be 2 cases :1st case when u seat all the boys 1st in 5 places then arrange the girls .2nd case you can seat the girls 1st then arrange the boys among them.

hence we take 2*

Hope its clear now.

Ashwin

Re: Permutation and Combination
by Saravanar B - Wednesday, 21 October 2009, 01:24 PM
  Hi TG Sir,

Thanks a lot for your article and I don't have any words to describe ur work sir...

I owe a lot... Great job sir... I have a great respect for you sir and long live TG familysmile

Have a small doubt sir :

1. Four friends go to a city in which there are 10 hotels. In how many ways can they stay ?

Sol :

(approach 1) : The first friend can stay in 10 ways, the second friend can stay in 10 ways, similarly the third and fourth..
                    
                   so the ans is 10*10*10*10 = 10000

(approach 2) : The first hotel can accomodate 4 friends or 3 friends or 2 friends or 1 friend or 0(zero). So nu of ways the 1st hotel can accomodate is 1+2+3+4 = 10 ( i have added 1,2,3,4 as it is either - "OR" case)

               Same goes for 2nd hotel, 3rd hotel and 4th hotel...
                    
               so the ans is 10*10*10*10 = 10000
*******************************************************
Now have a look at question no 2
*******************************************************
2. In how many ways can u post 10 letters in 4 letterboxes ?

Sol :

( approach 1 ) The first letter can be posted in 4 ways, similarly the second,third....tenth.
               So the ans is : 4*4*4...*4 = 4^10 = 1048576


(approach 2 ) : The first letter box can get '1' or '2' or '3' ..... '10' letters.So nu of ways
                the 1st letter can accomodate is 1+2+3+4+..+10 = 55

                Same goes for 2nd letter 3rd letter and 4th letter

                so the ans should be 55*55*55*55= 9150625

Doubt : I couldn't figure out what is wrong with my second approach.. Kindly explain me sir.. Also tell me given a question when to apply which approach...

My prob is I never get satisfied with one approach sir.. I always try out different approach..

I am sure ur explanation would be helpful to many of the students like me...

Thanks
Saran
Re: Permutation and Combination
by Total Gadha - Wednesday, 21 October 2009, 09:42 PM
  Hi Saravanar,

If first letter box posts even 1 letter or 2 letters etc. the second letter box does not have 10 ways.

Total Gadha
Re: Permutation and Combination
by skyskiers 44 - Monday, 26 October 2009, 02:55 AM
  Hello,

m not sure if this is the right section to post permutation problem, but anyways please help me out with the following

In how many ways can 10 soldiers stand in 2 rows such that there are 5 soldiers in each row?

Is the answer 7257600 or 7257600/2 ?

Puneet
Re: Permutation and Combination
by Nabanshu Bhattacharjee - Tuesday, 27 October 2009, 01:33 AM
  Hi TG

Thanks again,

Wanted to share the probabilistic approach to dearrangement problem with the TGites.

http://mathforum.org/library/drmath/view/56592.html

A query: If you go through the proof in the link I have given, you can see that they are considering 4 out of 5 letters placed in the correct envelope. How come there result matches with the one given in this thread? I mean i could not find where I am going wrong. Please help!!!

Re: Permutation and Combination
by Saravanar B - Friday, 30 October 2009, 12:26 AM
 

Hi TG,

Thanks for your reply...

Yes i got it... the second letter box need not have ten ways...

Is there a way to solve this question in the letter box perspective (approach 2) ?? I tried to solve but couldn't get the ans as well as the approach..sad

Saran

Re: Permutation and Combination
by swetha chinthareddy - Monday, 2 November 2009, 03:35 PM
  6 Balls of different colours are to be distributed among 2 boys. What is the probability that each boy gets same number of balls?

a) 5/16
b) 15/64
c) 1/2
d) 3/8
Re: Permutation and Combination
by nishchai nevrekar - Wednesday, 4 November 2009, 05:58 PM
  Hi,
I would like to point out the that the solution to the 5 BOY , 5 GIRL MOVIE movie problem with no girls sitting together is a little flawed...

It doesnt take in to consideration this arrangement
GBGBGBBGBG n many others lik this....

so the soln to this has to be considered by keeping objects (or people for this problem) who are not supposed to be together as dynamic and others as static...
here boys have to be static n girls dynamic... as shown below
_B_B_B_B_B_
so girls only have six positions to go into and also they will never be together as we limiting them by providing only one slot btw boys.
Hence refer to them as dynamic as they can take 6 positions.
so, 6P5 = 6x120
while boys who can be together are static cause they can only take 5 positions.
5P5 = 120

so, total no ways  = 120x120x6 = 14400x6

Hope this was helpful...
Also, Please correct me if I am wrong :D
Re: Permutation and Combination
by Total Gadha - Friday, 6 November 2009, 01:46 PM
  Hi Nishchal,

You forgot the condition mentioned in the question- every boy wants to sit with a girl.

Total Gadha
Re: Permutation and Combination
by saurabh prabhudessai - Friday, 6 November 2009, 03:42 PM
  in the condition mentioned by nischai every boy has at least one gal on his side!!
Re: Permutation and Combination
by nishchai nevrekar - Friday, 6 November 2009, 03:59 PM
  in these types of arrangements GB GB BG BG BG
Here every boy is sitting with one girl.... it the condition was
"every  boy wants to sit in-between 2 girls" then i would be a diff case altogether.
If not then plz temme... y these cases are not to be considered...
Re: Permutation and Combination
by vineet jain - Tuesday, 17 November 2009, 12:41 AM
  Hi TG!!
superb article
I have a question which has been asked by Rohit also.i am using same language as him

"In how many ways can 4 persons be seated out of 5 boys and 3 girls in 4 different seats?

While explaining you said it is not the simple formula based P&C problem.

Though, I think that it could be solved in very simple terms without breaking down the problem into cases(the way you already suggested as the concept).

Two aspects to this problem:

1. Selecting 4 persons from 8 persons(5 boys+3 girls) which can be done in 8C4 ways

2. Arranging the 4 persons on 4 different seats for each selection which can be done in 4! ways

So total number of ways = 8C4 * 4! = 70*24 = 1680 (same as 8P4)"

please correct me if i am doing it wrong
Re: Permutation and Combination
by SHAUNAK A - Tuesday, 24 November 2009, 08:16 PM
  Hello TG sir,

That's CAT like article on Perm Com......simplicity exemplified...

Thanx Sir...
Re: Permutation and Combination
by Harish A - Wednesday, 25 November 2009, 09:32 PM
 

Hi TG,

This is My first post in this is forum.

It is great disappointment I would like say, even after going through the Lessons multiple times, I am unable to solve even a single problem. It is not because of the way you explain things, but it is my ability.

I am sure with these given set of skills, I am ill equipped to tackle CAT. 

It may be of great help if you can let me where the solutions are present I can just read them for purpose of learning.

Thanks,
Harish.A
Re: Permutation and Combination
by aritreyee chaudhuri - Wednesday, 2 December 2009, 08:50 PM
 

Hi TG

I loved this article..thanks a lot for ur effortssmile..this helped me recollect many of the concepts learnt earlier..

but i am not clear about the 'painting faces of a cube' problem..two other guys Syd S and Anirban Bhar have also asked the same question that i am going to type now..

When we fix the color of one side of a cube..we have 6 possibilities (as there are six colors)..and in that case the opposite side will have 5 possibilities..for the remaining we can use the circular perrmutation rule..so the remaining sides can be painted in (4-1)! ways..

So shouldnt the number of ways in which the cube can be painted, be 6x5x3! ?

Can u please clarify this?

Re: Permutation and Combination
by gurtez singh - Thursday, 11 February 2010, 05:56 PM
  nice work sir ...........
                  i hve a prob...                                                                  

~  there r four sections of a paper wid a max. marks f 45 fr each section & to qualify 1 have get min. 90 marks then in hw many ways he cn qualify da exam??????????????
Re: Permutation and Combination
by vijayshree menon - Monday, 5 April 2010, 10:05 PM
 

Hi ,

Can any1 explain me this

In the above post , TG sir cited an eg

@ an election meeting 100 ppl address rally. Order followed PM then MP followed by MLA

Solution given: Let PM = P MP denoted as Q and MLA as R. 10 speakers adddress rally in 10! ways. this includes PQR PRQ QPR QRP RPQ RQP. Now Since need only PQR we divide 10! by 6.

I have a genuine prob here . First of all its decided tat PQR will be the order which means we need to keep PQR fixed and only count the remaining 7 ppl arrangement. Y cant it be so ??? I know I wrong where am I making the mistake ....

Im a real dumb one to have such doubts ... but want to make it clear  ....

Pls TG sir or Dagny Mam or who ever cam amswer this q

 Pls help Pls...

 

Re: Permutation and Combination
by Aspirend Achieve - Monday, 10 May 2010, 02:27 PM
 

Hi TG Sir,

can you please explain the solution for problem 3

circular -linera query
by robin catch - Friday, 25 June 2010, 03:41 PM
 

DEAR TG..help me on this??

 the number of ways that 4 girls and 5 boys can sit around a table st no two girls sit together ?

is there a general formula??

what would be the solution for a linear table, ie 4 girls and 5 boys( no 2 girls sit together)??

what would be the soln had 2 girls been always together, (case1:around a round table

                  case 2: in a linear arrangement)???

 

thnx in advance, highly appreciate for quick reply  

imlovinmath wink 

Answers
by Ramit Goyal - Monday, 28 June 2010, 11:04 PM
  1 a
2 b
3 d
4 c
5 c
6 a
7 d
8 c
9 c
10 b
11 b
12 c
13
14 a
15
16
17 b
18 d
19 a
20 c

Please review and notify me for unattempted questions and wrong answers
Re: Permutation and Combination
by sonali pawar - Thursday, 8 July 2010, 03:12 PM
 

Hi TG,

As usual beautiful article!!!!

Just one doubt......

In the part of Distribution in the problem of 5 balls and 3 boxes case IV i.e when all balls and boxes are dissimilar, for distribution 113 and 122 u say the no of distributions are 3!. Shouldn't it be 3!/2! = 3 as there are 2 groups with same number of things?

Please correct me if m wrong!

Re: Permutation and Combination
by TG Team - Thursday, 8 July 2010, 05:32 PM
 

Hi Sonali smile

Though in 113 and 122 distribution there are two groups with same number of things but still they are different groups because of different balls. That's why it is 3! in place of 3!/2!

Hope it is clear. smile

Re: Permutation and Combination
by Naman Mirchandani - Tuesday, 20 July 2010, 04:29 PM
 

Sir,

I have one confusion:

In how many ways can 5 similar balls be distributed in  3 different boxes?

Ans: (by method given here) 21

But there is one basic question in you article (like in how many can one post 10 letters in 4 letter boxes:     4 * 4 *4.....10 times : 4^10....).....

So if we apply the above method considering similar letters and different letterboxes .....answer will come : 13C3 ?

Have we considered all letters dissimilar to get 4^10? (can it be generalised where nothing is mentioned about similarity or dissimilarity ? )

 

Pls help ??????????? ....

Naman

Re: Permutation and Combination
by Siddharth Khanna - Saturday, 7 August 2010, 09:36 PM
  Hi TG Sir ,

(
in these types of arrangements GB GB BG BG BG
Here every boy is sitting with one girl.... it the condition was
"every  boy wants to sit in-between 2 girls" then i would be a diff case altogether.
If not then plz temme... y these cases are not to be considered...
)

Even I have the same doubt as mentioned by nishchai .
Cant we consider this arrangement .

 
Re: Permutation and Combination
by ravi baranwal - Sunday, 8 August 2010, 03:32 AM
  agree wid siddharth n nischai.
Re: Permutation and Combination
by kesav singh - Wednesday, 11 August 2010, 11:01 PM
 

hi everyone!

can anybody help me in below question?

 how many different sums can be formed with the following coins

5 rupee,1 rupee,50 paise, 25 paise, 10 paise, 3 paisa , 2 paisa & 1 paise?

plz eleborate

Re: Permutation and Combination
by Ravi Mathur - Tuesday, 17 August 2010, 09:16 PM
  There are 10 steps in a staircase and a person has to take those steps. At every step the person has got a choice of taking 1 step or 2 steps or 3 steps.The number of ways in whch person can take those steps is??

Can anyone please help me with this staircaseproblem.
Re: Permutation and Combination
by versha jhunjhunwala - Sunday, 29 August 2010, 12:05 AM
  Out of 21 tickets marked with numbers from 1 to 21 ,three are drawn at random, find the probability that the three numbers on them are in A.P

can u please explain the complete method?
Re: Permutation and Combination
by NITIN NIGAM - Tuesday, 31 August 2010, 07:57 AM
  Hi Versha...my approach is as follows:

from 1 to 21 find the no. of AP's that u can get and there will be cases accordingly:

1. AP with d = 1 , we have 21 nos. Among these 21 nos. we can choose 3 nos. which are in AP as 1,2,3 or 2,3,4 or 3,4,5 and so on...hence total no. of cases will be 19 here with last case ending at 19,20,21

2. AP with d = 2, we have two series here
a) 1,3,5,7......21 = 11 nos.
hence nos. can be chosen as 1,3,5 or 3,5,7 and so on ..so here TOTAL CASES = 9
b) 2,4,6,8......20 = 10 nos.
hence nos. can be chosen as 2,4,6 or 4,6,8 and so on..so here TOTAL CASES = 8

3. AP with d = 3, we have the following series
a) 1,4,7,10,13,16,19 -> total cases = 5
b) 2,5,8,11,14,17,20 -> total cases = 5
c) 3,6,9,12,15,18,21 -> total cases = 5

4. AP with d = 4, we have the following series
a) 1,5,9,13,17,21 -> total cases = 4
b) 2,6,10,14,18 -> total cases = 3
c) 3,7,11,15,19 -> total cases = 3
d) 4,8,12,16,20 -> total cases = 3

5. AP with d = 5, we have the following series
a) 1,6,11,16,21 -> Total cases = 3
b) 2,7,12,17 -> Total cases = 2
c) 3,8,13,18 -> Total cases = 2
d) 4,9,14,19 -> Total cases = 2
e) 5,10,15,20 -> Total cases = 2

6. AP with d = 6, we have the following series
a) 1,7,13,19 -> Total cases = 2
b) 2,8,14,20 -> Total cases = 2
c) 3,9,15,21 -> Total cases = 2
d) 4,10,16 -> Total cases = 1
e) 5,11,17 -> Total cases = 1
f) 6,12,18 -> Total cases = 1

7. AP with d = 7, we have the following series
a) 1,8,15 -> Total cases = 1
b) 2,9,16 -> Total cases = 1
c) 3,10,17 -> Total cases = 1
d) 4,11,18 -> Total cases = 1
e) 5,12,19 -> Total cases = 1
f) 6,13,20 -> Total cases = 1
g) 7,14,21 -> Total cases = 1

8. AP with d = 8, we have the following series
a) 1,9,17 -> Total cases = 1
b) 2,10,18 -> Total cases = 1
c) 3,11,19 -> Total cases = 1
d) 4,12,20 -> Total cases = 1
e) 5,13,21 -> Total cases = 1

9. AP with d = 9, we have the following series
a) 1,10,19 -> Total cases = 1
b) 2,11,20 -> Total cases = 1
c) 3,12,21 -> Total cases = 1

10 AP with d = 10, we have the following series
a) 1,11,21 -> Total cases = 1


now total no. of AP's cases that you can have from these nos. is
19+17+15+13+11+9+7+5+3+1 = 100

now total no. of ways of selecting 3 nos. from given 21 nos. are 21C3 = 1330

Hence the required probability is 100/1330

I could not think of a shorter approach but i certainly feel that there must be one....TGites please help here...kindly post a shorter or simpler approach if you find one...and let me know if the answer is correct...


thanks
Nitin

Re: Permutation and Combination
by deepti anand - Monday, 6 September 2010, 08:50 PM
  hii...hv a ques..plz help me out


set A is formed by selecting some of the numbers from the first 100 natural numbers such that the HCF of any two numbers in the set are same

Q1:if every pair of set A has to be realtively prime and set A has max number of elements possible,then in how many ways set A can be selected..??
1)64
2)96
3)72
4)108

Q2: if the HCF of any two numbers in set A is 3, then what is the maximum no of elements set A can have..??
1)10
2)12
3)11
4)14
Re: Permutation and Combination
by Animesh Devarshi - Saturday, 11 September 2010, 07:23 PM
 

Hi ,

reg question: how many ways can for persons be seated out of 5 boys and 3 girls on four different seats?

why can't we go for 8C4?

Re: Permutation and Combination
by harleen mann - Monday, 4 October 2010, 01:27 AM
  @deepti anand

Ans 1.
thinking part: here, for elements to be max, it can be understood that we need to take all primes numbers <100 and "1". therefore we will have 26 elements in that set.
A little more thought to it will get to the point that we can also select powers of the primes nos. (<100). so here goes a little paper work part:
1
2^1 to 2^6 (highest 6 since 2^7 >100)
3^1 to 3^4
5^1 to 5^2
7^1 to 7^2
11
13............
so we would have: 6*4*2*2 = 96 ways!!

Ans 2.
HCF = 3; this means that the only common factor between any 2 elements should be 3.
To make this possible, start taking numbers from 3 onwards multiplied with primes,
i.e. 3*1, 3*2, 3*3, 3*5....3*31.
These are a total of 12 numbers!!
Re: Permutation and Combination
by Black horse - Wednesday, 13 October 2010, 09:15 PM
  There cant be a better article on P&C than this . cool
Re: Permutation and Combination
by Pritesh Ranjan - Tuesday, 19 October 2010, 03:11 PM
  hi TG..
i had 2 quests..i was wondering if u cud plz help me out wid dem..

1.)How many 6 digit numbers contain exactly 4 different digits..??

2.)Six white and Six black balls of the same size r 2 be distributed among 10 urns such that each urn contains at least one ball.What is the total no of distributions..????
Re: Permutation and Combination
by Ankit Mittal - Friday, 29 October 2010, 02:27 PM
  Hi, I am asking this question to get my concepts right , here is the question :
Determine the number of 5 card combinations out of a deck of 52 cards if there
is exactly one ace in each combination.

two ways of solving this:
4C1 X 48C4

&

4(ace can be selected in 4 ways) X 48(next card can be selected in 48 ways) X 47(next card can be selected in 47 ways) X 46(..) X 45(..)



which one is the right way and why ?
Re: Permutation and Combination
by TG Team - Friday, 29 October 2010, 04:00 PM
 

Hi Ankit smile

First one is correct.

In the selection of five cards their arrangement doesn't matter but in your second method the four cards have been ordered and that needs to be divided by 4! which will again fetch the same result as calculated by first method. smile

Re: Permutation and Combination
by Ankit Mittal - Friday, 29 October 2010, 06:57 PM
  Thanks very much for clearing the doubt , I got it right now for this one. Actually I am always confused when to consider the order and when not.usually if arrangement is asked in the question I consider the order otherwise not.
Re: Permutation and Combination
by saanthwana T - Thursday, 3 February 2011, 06:37 PM
  This article really helped me to get all the basics of P & C

Thank you very much

It helped me get all the fundas.. so that I need not depend on 3 4 sources to get knowledge of single concept smile
Re: Permutation and Combination
by saanthwana T - Tuesday, 8 March 2011, 09:30 PM
  Hello,

I have small doubt

for permutations with similar objects we use the formula

n!/ (k1! k2! k3! ....)

where n things are to be arranged, among which
k1 is the number things of one type
k2 is the number of things of 2nd type
soon ..........

now when among the n things above, if only r things are to be selected and arranged then can we use the below formula?

nPr / (K1! K2!.....Ki!)
Re: Permutation and Combination
by Neelkamal Biswas - Thursday, 28 April 2011, 10:57 AM
  Hi TG,

In the solved example of 5 boys & 5 girls in a cinema, the two ways of sitting are given as :      BGBGBGBGBG & GBGBGBGBGB

but we are not taking into account the possibility of GBGBGBGBBG as even in this situation no two girls sit together.
The answer should then be : 5! * 6! as after arranging 5 boys there are six places in which the girls can be arranged
_B_B_B_B_B_

Please advise on this .

Re: Permutation and Combination
by TG Team - Thursday, 28 April 2011, 07:08 PM
 
Hi Neelkamal smile

In the given example: it should be the desire of the girls also that they won't let two boys sit together otherwise your version is correct. smile

Kamal Lohia
Re: Permutation and Combination
by deepti saluja - Monday, 2 May 2011, 12:16 PM
 

Hi Kamal,

Can u pls post the answers to the problems mentioned above (P n C), otherwise how do i check which 1 is right n which 1 is wrong. Pls help.

Thanks

Re: Permutation and Combination
by TG Team - Monday, 2 May 2011, 04:40 PM
 
Hi Deepti smile

We intentionally don't provide the answers/solutions to problems so that you can always relish the taste of problem solving. Also most of the problems have been discussed already many times in above posts. If you are unable to solve some question or have some doubt in a particular question, you are more than welcome to discuss. But please don't pressurize to provide answers to all the problems. smile

Kamal Lohia 
Re: Permutation and Combination
by deepti saluja - Tuesday, 3 May 2011, 05:53 PM
 

Hi Kamal,

Thanks for the quick reply.

Please help in these doubts:

Q4 license plates, i got 26^3 * 220 , what's the correct answer?

 please make me understand, i could not solve these.

Q13 4 persons be chosen from a row of 10 persons,st no two persons are sitting next to each other.

Q15 Indian n Japanese teams

Q26 3 blocks of 5 seats each,  20 seats in a row..

Thanks N Regards,

Deepti

Re: Permutation and Combination
by TG Team - Wednesday, 4 May 2011, 02:39 PM
 
Hi Deepti smile

4. Your answer is correct and surprisingly it's not given in the options. (Though we have already corrected in our assignments at TathaGat)

13. Lets make sit the remaining 6 persons of 10. Now we are to select 4 places only out of 7 available from where the chosen 4 persons were picked which can be easily done in C(7, 4) = 35 ways. Option (b) is correct.

15. We are to just determine the number of arrangements of I, I, I, I, J, J, J which is given by 7!/(4!)(3!) or C(7, 3) = 35. Option (a) is correct.

26. Just place those 3 blocks of 5 seats each, so that you have used now 15 seats and remaining 5 seats are to be distributed among the four gaps only. That's equivalent to find the number of whole number solution of a + b + c + d = 5 which is given by C(5 + 4 - 1, 4 - 1) = C(8, 3) = 56.

Hope this is clear. smile

Kamal Lohia     
Re: Permutation and Combination
by nikita dhanuka - Wednesday, 11 May 2011, 10:31 PM
  Hi Kamal

i have one nagging doubt in circular arrangements..

suppose if we have 6 distinct beads. then 5!/2 distinct necklaces can be formed.
how to go about a similar sum if some of the beads are identical.
the exact question is:

"A necklace is to be made using 6 red, 3 blue and 3 green beads such that, no two red beads are adjacent to each other. All the beads of the same colour are identical. In how many ways can this necklace be formed?"

the way i approached the sum...all the red beads can be arranged in only 1 way. this leaves 6 spaces between them. these spaces have to be filled with the remaining 3 blue and 3 green beads. now i am stuck..this can be done in (5!/2*3!*3!)ways or (6!/2*3!*3!)ways??
please explain..

Re: Permutation and Combination
by TG Team - Friday, 13 May 2011, 03:32 PM
 
Hi Nikita smile

In the numerator part there will be surely 5! and not 6! but why are you putting a 2 in denominator. Kindly express your thoughts.

Kamal Lohia
Re: Permutation and Combination
by destiny unruled - Friday, 13 May 2011, 10:30 PM
  Hi Nikita smile

In case of circular permutations if we have identical things then it gets very complicated.

For the time being lets not consider the necklace case and just consider the permutations of 6 red, 3 blue, and 3 green beads around a circle such that no two red beads are adjacent to each other.

Luckily here we just have to arrange the blue and green beads between red beads as no two red beads can be together.

So, now the problem reduces to permuting 3 blue and 3 green beads around a circle.

Hence, answer should be 5!/(3!*3!), but its not an integer.

That means something is wrong.

So, lets go back to basics. Why do we divide by n while counting circular permutations to get (n - 1)! permutations.

Because we can get n different permutations by rotating a particular arrangement, but actually they are same. Thats why we divide by n. This is the case when we have all different objects(here we have "rotational period of n"*).

* - Rotational period of n means that if you rotate it n times it goes to an indistinguishable state from the first one

Now, suppose we have 3 red balls and 3 black balls. Now here we can have cases when we have rotational period of 1 or 2 or 3 or 6 (factors of 6).
But 1 is not possible coz then we should have all balls of same color.
Also 3 is not possible as then we need to have equal groups of three balls which means by Pigeonhole there must be at least 2 balls of some color
Hence at least 4 balls of this color in all, which is not possible since there must be 3 of each color. Hence, only possibilities are 2 or 6

BGBGBG has a rotational period of 2, while all other cases have rotational period of 6.

Say there are x cases of rotational period of 6, then
6*x + 2*1 = C(6, 3) = 20
=> x = 3

=> There will be just 4 distinct permutations of 3 red balls and 3 black balls around a circle.

So, if you wants to solve a similar kind of question, then just find the number of different permutations of 6 black chairs and 6 white chairs around a circle.
Re: Permutation and Combination
by destiny unruled - Friday, 13 May 2011, 10:51 PM
  Sorry, I forgot to solve it for necklace.

In case of necklace, BBGBGG and GGBGBB will also look alike, as now we can look at the necklace from the back side also.

So, only 3 possible permutations.
Re: Permutation and Combination
by tgdel156 tg - Sunday, 22 May 2011, 11:57 PM
 

Hello Sir !

Please let me know the solution to questions involving the equation

like

1) 3x+ y + z= 40  where x,y,z are positive integer

2) x+y+z = 10 if x, y, z are distince positive integers.

3) x+y+z>= 10 ===> x+y+z -k =10

 

Thank You

Gaurav

Re: Permutation and Combination
by nikita dhanuka - Wednesday, 25 May 2011, 05:46 PM
  hi kamalsmile

i have divided by 2 as the necklace is inanimate and can be flipped over. so, for every 2 different circular arrangements, we actually have only one necklace. hence, divided the total no. of possible circular arrangements by 2 to get the no. of distinct necklaces.
even i was inclined towards 5!/3!*3!*2 as the answer. but, this is obviously wrong as it doesn't give an integer value.
the book gives the answer as 6!/3!*3!*2. i am not being able to understand the logic behind this.
i will appreciate if you can discuss your approach for this sum. and also if you can shed some light on these kind of sums in general...circular arrangement of things, some of which are identical.
thanks,
nikita
Re: Permutation and Combination
by nikita dhanuka - Wednesday, 25 May 2011, 05:56 PM
  Hi Destiny smile

thanks for taking the time out. the book gives the answer as 6!/3!*3!*2 which works out to be 10.
i am still pretty confused in this regard. thoughtful
Re: Permutation and Combination
by destiny unruled - Thursday, 26 May 2011, 12:56 AM
  Hi Nikita smile

I'm pretty sure that the answer is 3 and not 10.

Did you get the logic I tried to explain in my last post???
Re: Permutation and Combination
by TG Team - Thursday, 26 May 2011, 05:36 PM
 
Hi Gaurav smile

1. x varies from 1 to 12 and number of solutions for x and y in each case form an AP. You just need to add them to get the total number of solutions. It comes out to be 3(1 + 2 + 3 + ... + 12) = 234.

2. You can easily count the ordered triplets by assigning values to x as 1, 2, ..., 7 - that comes out to be 24.
OR just count the ordered triplets when all the three variables are positive integers i.e. C(9, 2) = 36. And subtract the cases when any two of the variables are same (e.g. 11.., 22.., 33.., 44..). So removing 3*4 = 12 ordered triplts from total 36, we are left with 24 ordered triplets of three distinct positive integers which add up to 10.

3. I am sure that there is some error in question because for the stated question answer is going to be infinitely large. See If x + y + z 10, that means the solutions when x + y + z = 10 will be added to the number of solutions of x + y + z = 11,...and so on. So there is no end and the answer is (infinity) for sure. smile

Kamal Lohia
  
Re: Permutation and Combination
by Nikunj Bajaj - Sunday, 21 August 2011, 02:56 PM
  I have a doubt:-

In the question where we need to paint 6 faces of the cube with 6 different colors, why have we not multiplied the outcome with 6?

when we fixed the first color, it could have been done in 6 ways..

what m i missing?
Re: Permutation and Combination
by sowmya k - Monday, 22 August 2011, 11:39 PM
  Hi TG

came across this forum damn too late sad but simply brilliant are your lessons and tips... as much about basics and foundation as they are about tricks and shortcuts minus any heavy duty formulae.. a lot of my friends too are hooked onto this site now enroute me ofcourse ;)
Re: Permutation and Combination
by anupam chaturvedi - Wednesday, 24 August 2011, 03:07 AM
 
Hi TG,

Its a wonderful article.This is the first time I have been able to understand P&C, that fear has somewhat vanished now..!! smile

Still there are quiet a few things that I couldn't understand. mixed

1) No. of ways of distributing 'n' similar objects into 'r' similar groups.

2) No. of ways of distributing 'n' distinct objects into 'r' similar groups.

and a few questions that I'll post soon.

Looking ahead for your reply on above 2 problems!

Thanks a ton in adv. ! smile
Re: Permutation and Combination
by ketav sharma - Wednesday, 7 September 2011, 12:07 AM
  come on man solved all questions where is the solution sad

I was highly disappointed after seeing no key
Re: Permutation and Combination
by priyanka j - Wednesday, 7 September 2011, 11:50 AM
  Sir,

1) Is d ans of "god boy bad boy movie" ques is 8*6! =5760

2) In d next ques of "election meeting" . How wud we know how many MP,MLA are there out of 10 speaker. Not able to understand d sol. Plz explain.

3)IN d ques first 0 precedes first 1,precedes d first 2. Does it mean frst 0 pecedes frst 1&2 or it means frst 0 preceds frst 1& first 1 preceds frst 2.

Please help me out.

Thankssmile
Re: Permutation and Combination
by priyanka j - Wednesday, 7 September 2011, 11:53 AM
  ketav ,
u can get ans here

http://totalgadha.com/mod/forum/discuss.php?d=3683
Re: Permutation and Combination
by amit kumar - Thursday, 29 September 2011, 07:37 PM
 

Dear TG Sir/kamal Sir,

What is the answer to this problem

In how many ways can we put 5 different balls in 4 different boxes so that each box should have atleast one ball?

My answer is 240.

regards

Re: Permutation and Combination
by amit kumar - Thursday, 29 September 2011, 08:24 PM
 

Dear TG sir/Kamal Sir,

The article is awesome!!!

Can you please provide the solution to Q 23(c), Q 27, Q 28.

Can you please confirm the answer to Q 29, my answer is 2^20.

Regards,

 

Re: Permutation and Combination
by krishnadev kashid - Thursday, 29 September 2011, 09:06 PM
  very good elaboration of concepts..thanks a lot....!! thoughtful thoughtful
Re: Permutation and Combination
by TG Team - Friday, 30 September 2011, 11:48 AM
 

Hi Amit smile

Please post the question statements here so that it becomes easier for everyone to try their hands at the problems.

Kamal Lohia

Re: Permutation and Combination
by amit kumar - Friday, 30 September 2011, 04:38 PM
 

Dear Kamal Sir,

Please answer the following queries:

Q.23(c) In how many ways can the letters in UNUSUAL be arranged such that they have no consequtive Us?

My answer is 240. Can you please confirm?

Q.27 There are 50 juniors and 50 seniors. Each class has 25 men and 25 women. In how many ways can an 8 person committee be chosen so that it includes 4 women and 3 juniors?

Please solve this question.

Q.28 In how many ways can 2n people be divided in n pairs?

My answer is (2n)!/(n!(2^n)). Can you please confirm?

Q. 29 I have two checkout registers, and twenty customers. What formula will find how many different ways I can arrange them? Order does matter.

My answer is 2^20. Can you please confirm?

Q. 36 If combinations of letters be formed by taking 5 letters at a time out of the letters of the word "METAPHYSICS", in how many of them will T letter occur?

My answer is 210. Can you please confirm?

Q.37 How many sequences a1, a2, a3, a4, a5 satisfying a1<a2<a3<a4<a5 can be formed if ai must be chosen from the set {1,2,3,4,5,6,7,8,9}?

My answer is 126. Can you please confirm?

Q. In how many ways can we put 5 different balls in 4 different boxes so that each box should have atleast one ball?

My answer is 240. Can you please confirm?

Thanks & Regards,

Amit

Re: Permutation and Combination
by TG Team - Friday, 30 September 2011, 05:34 PM
 

Hi Amit smile

Q.23(c) In how many ways can the letters in UNUSUAL be arranged such that they have no consequtive Us?

Look after placing 3 Us we have 4 boxes (_U_U_U_) which must be containing the other four different letters with the condition that the middle two boxes have atleast one letter each. So we need to find the whole number ways of distributing 2 letters among 4 boxes i.e. C(5, 3) = 10 ways. As all the four letters are different so they can be arranged among themselves in 4! = 24 ways. So the final number of ways = 240.

Q.27 There are 50 juniors and 50 seniors. Each class has 25 men and 25 women. In how many ways can an 8 person committee be chosen so that it includes 4 women and 3 juniors?

We have 50 Juniors (25M + 25W) and 50 Seniors (25M + 25W) and need to select a team of 8 person which includes 4 women and 3 juniors (atleast or exactly is not mentioned). Assuming exactly, then we have the following cases:

4W + 4M from 3J + 5S

J - 3W and S - 4M + 1W - C(25, 3)×C(25, 4)×C(25, 1)

J - 2W + 1M and S - 3M + 2W - C(25, 2)×C(25, 1)×C(25, 3)×C(25, 2)

J - 1W + 2M and S - 2M + 3W - C(25, 2)×C(25, 1)×C(25, 3)×C(25, 2)

J - 3M and S - 1M + 4W - C(25, 3)×C(25, 4)×C(25, 1) 

Q.28 In how many ways can 2n people be divided in n pairs?

It has been clearly taught in TG's lesson that required number of ways are (2n)!/(n!×2ⁿ).

Q. 29 I have two checkout registers, and twenty customers. What formula will find how many different ways I can arrange them? Order does matter.

First customer has 2 choices, next has 3 choices, next has 4 choices and the 20th have 21 choices. So answer is 21!

Rest later. smile

Kamal Lohia

Re: Permutation and Combination
by amit kumar - Friday, 30 September 2011, 06:19 PM
 

Dear Kamal Sir,

THANK YOUsmile for the detailed and lucid explanation. By the way can you please discuss the remaining possibilities of Q27.

Waiting for the rest.

Thanks & Regards,

Amit

Re: Permutation and Combination
by amit gupta - Friday, 30 September 2011, 08:05 PM
  In the question: how many four letters can be formed from the word Mathemetics why are the sub-dividing the problems into groups and then solving ? Can't we directly use 11P4 ??
Also, similar approach is used for the 3 students 3 city question ?
would like to know the reason for it ?
Re: Permutation and Combination
by amit kumar - Sunday, 2 October 2011, 11:09 PM
 

Dear Kamal Sir,

Please let me know the answers for the remaining questions:

Q. 36 If combinations of letters be formed by taking 5 letters at a time out of the letters of the word "METAPHYSICS", in how many of them will T letter occur?

My answer is 210. Can you please confirm?

Q.37 How many sequences a1, a2, a3, a4, a5 satisfying a1<a2<a3<a4<a5 can be formed if ai must be chosen from the set {1,2,3,4,5,6,7,8,9}?

My answer is 126. Can you please confirm?

Q. In how many ways can we put 5 different balls in 4 different boxes so that each box should have atleast one ball?

My answer is 240. Can you please confirm?

Thanks & Regards,

Amit

Re: Permutation and Combination
by TG Team - Monday, 3 October 2011, 01:46 PM
 

Hi Amit smile

What are the remaining possibilities for Q 27? Check.

Kamal Lohia

Re: Permutation and Combination
by TG Team - Monday, 3 October 2011, 04:53 PM
 

Hi Amit smile

Q. 36 If combinations of letters be formed by taking 5 letters at a time out of the letters of the word "METAPHYSICS", in how many of them will T letter occur?

Here we need to find out the number of cases when we are selecting a T for sure while selecting 5 letters off the word METAPHYSICS.

So other than T we must be selecting 4 more letters and we just need to find the number of ways of selecting 4 more letters. Now there are two cases: when all the four letters are different i.e. C(9, 4) = 126 ways and when two letters are same (i.e. both are S) and other two are different i.e. C(8, 2) = 28 ways.

So total cases must be 126 + 28 = 154 ways.

 

Q.37 How many sequences a1, a2, a3, a4, a5 satisfying a1<a2<a3<a4<a5 can be formed if ai must be chosen from the set {1,2,3,4,5,6,7,8,9}?

It is simply C(9, 5) = 126 ways.

Because in C(9, 5) ways you select five different numbers from the available 9 and arrange them(according to desired order) in only one possible way.

 

Q. In how many ways can we put 5 different balls in 4 different boxes so that each box should have atleast one ball?

There will be one box(which can be selected in 4 ways) containing two balls(which can be selected in C(5, 2) = 10 ways) and the remaining three balls can be arranged in three different boxes in 3! = 6 ways.

So required number of ways are = 4 × 10 × 6 = 240.

 

Kamal Lohia

Re: Permutation and Combination
by amit kumar - Tuesday, 4 October 2011, 12:22 AM
 

Thank you Kamal sir for all your explainations. your way of explaination is awesome....simple and clear!!!

I really appreciate your help.

Regards,

Amit

Re: Permutation and Combination
by TG Team - Tuesday, 4 October 2011, 01:24 PM
 

Hi Amit Gupta smile

The problem is of repetitive counting. Try to check all the cases and find the repetition.

Kamal Lohia

Re: Permutation and Combination
by Mohit Sharma - Wednesday, 9 May 2012, 01:05 PM
  Hi Kamal Sir,
Can you please explain the following two questions as I'm unable to solve them,

Q1) From 4 gentlemen and 4 Ladies a committee of 5 is to be formed. Committee consists of a president, a vice-president and three secretaries. Find the number of ways of selecting the committee with a maximum of 2 women and having at the maximum one woman holding one of the two posts on the committee.
a) 16
b) 512
c) 608
d) 256
e) 324

Q2) The crew of an 8 member rowing team is to be selected from 12 men, of which 3 must row on one side only and 2 must row on the other side only. Find the number of ways of arranging the crew with 4 members on each side.
a) 40320
b) 30240
c) 60840
d) 10080
e) None of these

Please help me sir.
Re: Permutation and Combination
by TG Team - Wednesday, 9 May 2012, 04:44 PM
 

Hi Mohit smile

Most important part is to get the correct interpretation of the question and then framing the solution.

1. What I interpret from the given information is that you are to choose 5 members from 4 M and 4W to form a committee. Of these five members one is a president, one vice-president and other three are secretaries.

Now conditions mentioned are there should be maximum of 2W i.e. cases can be 4M1W or 3M2W. Another condition is that maximum one women holds one of the two posts. Here what i interpret from the two posts is post of president/vice-president. That means cases will be that no woman is president or vice-president AND exactly one woman is president or vice-president.

Now computing the first case: 4M 1W - (i) no woman is president/vice-president = C(4, 1) × 4 × 3 = 48

(ii) one woman is president/vice-president = C(4, 1) × 2 × C(4, 1) = 32

Second case: 3M 2W - (i) no woman is president/vice-president = C(4, 3) × C(4, 2) × C(3, 1) × 2 = 144

(ii) one woman is president/vice-president = C(4, 3) × C(4, 2) × C(2, 1) × C(3, 1) × 2 = 288

So total cases are = 48 + 32 + 144 + 288 = 512

 

2. Again here I assume that these 3, which must row on one side and 2, which must row on other side are to be chosen certainly. That means I just need to choose 3 more out of 7 remaining in C(7, 3) ways. Now out of the three newly selected members one should be sent to 3 together's and remaining to other side with 2 together's which can be done in C(3, 1). Now these groups can be arranged among themselves in 4! each.

So total cases are = C(7, 3) × C(3, 1) × 4! × 4! = 60480.

I hope I haven't created any conceptual/calculational error. smile

Kamal Lohia 

 

Re: Permutation and Combination
by Mohit Sharma - Thursday, 10 May 2012, 10:43 AM
  thank you kamal sir,
 I really didn't understood the question... thanks a lot... smile
Re: Permutation and Combination
by Prerna Golani - Tuesday, 15 May 2012, 03:52 PM
  Hi TG
In how many ways six different faces of cube be painted in different color

why cant answer be 6*5*4*3*2*1
Re: Permutation and Combination
by TG Team - Tuesday, 15 May 2012, 05:37 PM
 

Hi Prerna smile

It would have happened if all the faces of cubes are different or differently named. But that is not the case, so you need to remove the identical ones.

Kamal Lohia

Re: Permutation and Combination
by Prerna Golani - Sunday, 20 May 2012, 04:05 PM
  but it is written six different faces
Re: Permutation and Combination
by neha aggarwal - Thursday, 31 May 2012, 02:16 PM
 
Hi Kamal Sir smile

Q.23(c) In how many ways can the letters in UNUSUAL be arranged such that they have no consequtive Us?

Look after placing 3 Us we have 4 boxes (_U_U_U_) which must be containing the other four different letters with the condition that the middle two boxes have atleast one letter each. So we need to find the whole number ways of distributing 2 letters among 4 boxes i.e. C(5, 3) = 10 ways. As all the four letters are different so they can be arranged among themselves in 4! = 24 ways. So the final number of ways = 240.


In above explanation i did not get the condition? And solution to it.. As in how did U get C(5,3)?

Please explain

Re: Permutation and Combination
by TG Team - Thursday, 31 May 2012, 02:45 PM
 

Hi Neha smile

If you didn't get that way, no probs. Here is another. smile

See, that no two U's are to to be together. So first arrange the remaining four letters (i.e. N, S, A and L) in 4! = 24 ways. Now between these four letters (including at ends) we have exactly five places at which U's can be placed.

Remember that at no place two U's can be together. So we need to select exactly 3 places out of these 5 in C(5, 3) = 10 ways such that no two U's are together.

Hence the total number of ways is = 24 × 10 = 240. smile

Kamal Lohia 

Re: Permutation and Combination
by Sowmyanarayanan P - Friday, 1 June 2012, 01:27 PM
 

Clear explanation sir, thanks..

Re: Permutation and Combination
by manik bhardwaj - Monday, 4 June 2012, 11:28 PM
  Hello Kamal Sir,

Please hep me with below mentioned questions:

In a class of 12 students , 7 boys and 5 girls.The class has 4 sessions each day,one each of arithmetic ,algebra,geometry and probability.These classes are to be held one after the other in 4 distinct time slots and can be in any sequence .Further there are 2 teachers available and they can teach any topic.

1.  if the sequence of class has to be arithmetic,algebra,geometry and probability,then what is number of distinct ways in which the session can be planned for entire week.

2.
In how many distinct ways can a session be scheduled for two consecutive days if both the teachers have to have equal no. of sessions..?
my take(24* 2^4)*(24* 2^4)

3. the entire class of 12 students is divided equally into two dofferent divisions and sessiosn are schedules for these two divisions 2ith classes being held simultaneously in 2 classrooms..in how many ways can session be planned for these two divisions for particular day?



Re: Permutation and Combination
by TG Team - Wednesday, 6 June 2012, 02:04 PM
 

Hi Manik smile

1. As sequence of class is fixed, so we are now to arrange for the teachers only at a paricular time slot. Now in one day, at any time slot, the teacher can be assigned in 2 ways. So number of ways of assigning teacher in one day is = 24 ways and for the entire week is = 228 ways.

2. When both the teachers have to have equal number of sesions on two consecutive days i.e. we have eight sessions which must be assigned to the two teachers equally i.e. 4 sessions each which can be achieved in C(8, 4) ways. Also the topics can be scheduled in the two days in 4! ways each. So the required number of ways are = (4!)² × C(8, 4) = 8!

3. Now we are to schedule two parallel classrooms and topics can be arranged for both of them in 4! ways each and teachers can be assigned to both in 24 ways in total. Hence total number of ways required are = (4!)² × 24.

Remember that in all these questions, a session referred to a particular topic arranged at a particular time slot taught by a particular teacher only. Students don't matter in schedulling of the session.

Kamal Lohia 

Re: Permutation and Combination
by manik bhardwaj - Wednesday, 6 June 2012, 02:35 PM
  Thanks alot Kamal Sir.
My another doubt in this topis is, like for above question mentioned by neha:

In how many ways can the letters in UNUSUAL be arranged such that they have no consequtive Us?

I have one confusion for the solution given by u, like here _n_s_a_l_, no two u's are together , so obvious answer is 4!* C(5,3)=24*10=240.

But, as per question no two u's can be together, but NSAL can be together, like _NS_A_L_,here answer would b like-4!*C(4,2)=24*6=144.

My take for this question is like, to normal permutation with all 7 letters- 7!/3!
Then no of ways when all U's are together, 5!=120
therefor,when no twou's are together= 840-120=720,

Please correct me if i am wrong..
Re: Permutation and Combination
by arsh arora - Thursday, 7 June 2012, 05:38 AM
 

hi manik,

   see you are including the cases when only two u's are together,which are to be discarded,there's the flaw!! as u had just debarred the cases in which all u are together and in 5c3 approach u will see the case which u had mentioned is already incorporated,as we are selecting the 3 blanks out of 5 blanks,hope its clear now

Re: Permutation and Combination
by manik bhardwaj - Thursday, 7 June 2012, 01:11 PM
  Hi arsh,

well Iam including all the three u's together,perhaps there are total 7 letters, 3 taken together which are to be counted as one and the rest 4 letters, so the in all the sum comes to 5.
Re: Permutation and Combination
by neha aggarwal - Thursday, 21 June 2012, 11:42 PM
  Hi TG Sir/Kamal Sir,smile

How should we approach such questions? :

x+y+z = 120. x,y,z <60 number="" of="" integral="" />
This is a type.. I really get confused in identifying the cases.. Hope am making sense.. Please explain the approach..

Thanks,
Neha
Thanks to all the authors nd contributing folks
by Kanwar Singh - Thursday, 5 July 2012, 02:45 PM
 

great work ......

But just one SHIKAYAT .....

We want more of these great articles

Please find some time !!!!

 

Re: Permutation and Combination
by Akash Kumar - Wednesday, 29 August 2012, 04:23 PM
  I think the answer will be

720*2*2*2

10 are there in total...1 couple will sit at one corner and other at another corner so 720 arrangement

2 couples can also interchange seats among themselves in 2 ways

Couple can also changes seats in 2 ways as 2 couple are there 2*2

Please correct if i am wrong
Re: Permutation and Combination
by poora gadha - Friday, 31 August 2012, 04:59 AM
  In the que where boys and girls go for spiderman movie
why cant be there an arrangment gbbgbgbgbg
still no girl is sitting together
Re: Permutation and Combination
by Alo Sanyal - Tuesday, 4 September 2012, 12:55 PM
  Hi TG,

This is a really fantastic article! smile I'd really appreciate some more help. Can you please explain how to find out the rank of a string when repetition of characters is allowed?

For instance, if I have the letters R, D, S, B and repetition of the letters is allowed, then the number of 4-letter strings = 4X4X4X4 = 256. So how do I find out what is the 110th string?

Regards,
Alo

Re: Permutation and Combination
by rimmi rimmi - Friday, 21 September 2012, 11:22 PM
  in the question of multiple choice test with 5 choices, i think there are 6 possible ways of answering the questions instead of 5 as a person can also leave the question/s unanswered which i think is the 6th way of doing a question.

therefore solution will be... 6^30

but i am not sure, so please let me confirm that my approach is correct or not..smile
Re: Permutation and Combination
by Rishabh singla - Sunday, 23 September 2012, 11:25 AM
  Great article...thanks a ton TG...your articles motivate to study more as concepts seems to be so simple....
I have a question , if you or anyone can help me out...
In how many ways 25 balls can be selected out of 15 identical red balls,20 identical blue balls and 25 identical green balls ?

My Approach : consider 3 persons A,B,C to whom I have to distribute these balls.So, I distribute all balls , say 15 Red to A , 20 Blue to B  and 25 Green to C .Total =60 .
Now I can take out 35 out of them as we want only 25 balls in total.
so, finding solution of equation x+y+z= 35 should give answer . But my answer coming wrong. Correct answer is 281. Can anyone tell me where I am going wrong.
Re: Permutation and Combination
by Sumit Anand - Monday, 24 September 2012, 12:11 AM
  Here,
r+b+g=25
total ways=27c2=351
Now
r
Re: Permutation and Combination
by Rishabh singla - Wednesday, 26 September 2012, 01:27 AM
  Hi Sumit,
 thanks for replying but I didnt got you.. where you got 27C2 from ?
Re: Permutation and Combination
by scrooze scrooze - Wednesday, 26 September 2012, 06:06 PM
  Awesome article..no words..
Re: Permutation and Combination
by dimpu n - Saturday, 29 September 2012, 10:49 PM
  In how many ways can 4 persons be seated out of 5 boys and 3 girls on four different seats

In TG's article he took different cases of taking

1) 4boys 2) 3boys 1 girl 3) 2boys 2 girls 4)1boy 3girls

and calculated the answer as 1680

there is no difference to place a girl or a boy

in first seat we can place any of the 8 persons

in second seat we can place any of the 7persons

in third seat any 6 and in fourth 5

ans : 8*7*6*5 = 1680

I got the same answer but i am just wondering if the approach is right are we need to take different cases.
Re: Permutation and Combination
by munjal upadhyay - Saturday, 6 October 2012, 06:05 PM
  first thank you for the wonderfull article .. I have a doubt that ,, in the example of how many even digits are that which are even and have no repeted digits .. Question : how do I reliaze that I need to take two case here,, if I am in exam hall ,I don't think that I get the strike in my mind .. if there is any trick for this then please deliver to me .. thank you ...
Permutation and Combination
by Rahul Agarwal - Tuesday, 23 October 2012, 06:21 PM
  There are 20 seats on a circular table 10 on either side. 20 persons have to be seated such that 5 particular people want to sit on 1 side and 4 particular people sit on the other side. what are the total number of ways?
Please answer this question
Re: Permutation and Combination
by The Dumbest - Wednesday, 3 April 2013, 07:05 PM
 

Hi sir,

Forgive me for being naive..thr is an example where there are computer user passwords where a password could be 6,7 or 8 char long...can this ques not be attempted like this...

lets take the case for a password with 6 char..now one of these char has to be a digit so there are 10 ways...and for the rest five char it could be a capital alphabet or a digit(26 + 10 ways for every char)..so 10 * 36^5....

similarly add to this the results for cases whr passwords are 7 and 8 char long..?

Re: Permutation and Combination
by rahul kankariya - Thursday, 27 June 2013, 07:44 PM
  how many arrangements of 6 0's, 5 1's and 4 2's are there in which
1)the first 0 precedes the first 1?
2)the first 0 precedes the first 1,precedes the first 2?
3)the first 0 precedes the first 2?

what is the answer of last two questions??
and can u plz guide me with the solution..
Re: Permutation and Combination
by Sankarshan Sridharan - Wednesday, 17 July 2013, 02:35 PM
  Hi TG,

I am a big fan of your articles. They are lucid and very informative. Recently I came across a problem which led me to another problem. This is what I want.

I have six alphabets a,b,c,d,e,f with the following relationship

a > b (pair)
c > d (pair)
e > f (pair)

Now how many permutations/arrangements of a,b,c,d,e,f are possible such that the higher number of the "pair" always precedes the lower number of the "pair".

For example acbdef is a permutation, ecfdab is another permutation. How many permutations will preserve this relationship?

Warm Regards,
Shanky
Re: Permutation and Combination
by Sankarshan Sridharan - Thursday, 18 July 2013, 06:00 PM
  Hi All,

Got my question answered through another forum. Solution is elegant. There are 720 different ways of arranging a,b,c,d,e,f.
In exactly half of these a will be before b. Call this SET X.In exactly half of SET X c will be before d and this be SET Y. In exactly half of SET Y e will be before f. So 720/(2^3) = 90 is the solution.

Warm Regards,
Shanky
Re: Permutation and Combination
by Dushyant Agarwal - Monday, 22 July 2013, 04:00 PM
  Under 'Permutation' section:

Q- In how many ways can you seat n boys in r seats ?

In this question you have mentioned that it can be done in nPr ways. However my doubt is what if n < r ? Do I need to take it (n >r) as an assumption ? Or else it would be rPn.

I believe, I am missing on some concept. Can you please clarify my doubt ?

Re: Permutation and Combination
by Rachit Pande - Saturday, 3 May 2014, 11:28 AM
  Hi TG,

For The example with 5 boys and 5 girls we need to satisfy the conditions that each boy sits with a girl and no two girls sit together. The question does not ask to make them sit alternatively. So, I feel that the case for GBGBGBGBBG is also possible and similar other cases. so in total if we first fix 5 boys in 5! ways then we hav 6 place to fix girls which can be done in 6P5 ways thus total count comes to 5!x6!=86400

Plz tell me if I am wrong.

Rachit smile
Re: Permutation and Combination
by TG Team - Tuesday, 13 May 2014, 01:37 PM
  Hi Rachit smile

Your answer and approach is correct. I'll fix it there. smile

Kamal Lohia
Re: Permutation and Combination
by Aakash Gupta - Sunday, 8 June 2014, 11:23 PM
  Hello TG,

In the kind of problem with unit coefficient--
  a1+a2+a3..... + ar = n (constant),
             the number of solutions are =  C(n+r-1, r-1) in case  a1, a2.. ar >= 0 &           the number of solutions are = C(n-1, r-1) in case    a1, a2.. ar >= 1

there's no upper limit on variable on left side here.

But what if there's an upper limit on variable like-


Question: A dice is rolled 3 times. What are the number of ways in which the sum of numbers appearing on the dice is 12?

Here, let the outcomes be a,b and c.
therefore, a + b+ c = 12, where 1<=a,b,c <= 6


Re: Permutation and Combination
by sravani jaganti - Monday, 23 June 2014, 11:19 PM
  sir can you please give us article on permutations around regular and irregular polygons. i,e the number of waysof arranging people around a table of 'n' sides , around a rectangle.such kind of permutations.please reply soon as possible sir.
Re: Permutation and Combination
by TG Team - Tuesday, 24 June 2014, 02:30 PM
  Hi Aakash smile

Maximum value which a, b, c can get is 6, so let's give each of them a 6. But in this distribution I have given a total of 18 while available quantity was 12 only. Right?

So I need to take total of 6 back from the three.

Let's say I take back a' from a, b' from b and c' from c such that
a' + b' + c' = 6 where a', b', c' are whole numbers but none of them can be zero. (Think why?)

It's easy now to calculate this. Try and revert if you face any difficulty.

My answer to this one is: C(6 + 3 - 1, 3 -1) - 3 = C(8, 2) - 3 = 28 - 3 = 25.

Kamal Lohia
Re: Permutation and Combination
by TG Team - Tuesday, 24 June 2014, 02:33 PM
  Hi Sravani smile

Above article deals with all the varieties which you are asking for. If you any particular query, then please post your question so that everyone can try their hands at it.

Kamal Lohia
Re: Permutation and Combination
by Abheek Das - Sunday, 3 August 2014, 10:55 AM
  (3^1/5 +7^1/8)^100
How many terms in the expansion are integers?
Re: Permutation and Combination
by TG Team - Monday, 4 August 2014, 12:58 PM
  Hi Abheek smile

You only need to find integral solution of the equation 5x + 8y = 100.

And I am sure that you can do it easily. smile

Kamal Lohia
Re: Permutation and Combination
by Abheek Das - Monday, 4 August 2014, 01:23 PM
  My approach for finding the integral solution to the equation:
5x+8y=100
100 / product of 5x8
I can take it as the product since 5 & 8 are co-prime.
This gives me roughly 100/40=2. something
So the number of integral solutions could be either 2 or 3.

Second step, (100-5x)/8=y
x=4, y =10
Now x will increase by the coefficient of y and y will decrease by the coefficient of x
The possible values are
x=4, y=10
x=12,y=5
x=20,y=0

Therefore, the number of integral solutions for the equation is 3.

Question: Does this method work in all cases or is it intrinsic to the problem above also should this be the ideal method for dealing with questions which require you to find the integral solutions
Re: Permutation and Combination
by TG Team - Monday, 4 August 2014, 01:28 PM
  Both methods are perfect and are applicable everywhere. Still second one will give you exact number of solutions while first one only give a hint for the answer.

Kamal Lohia
Re: Permutation and Combination
by Dimple Mehta - Monday, 3 November 2014, 03:30 PM
  hi TG
thnx for such a wonderful article...please tell me if u posted any article on Integral Solution?? I couldn't find dat..pls hlp me in dat concept..
thnx
Re: Permutation and Combination
by Madhuraj Tripathi - Saturday, 21 March 2015, 09:09 PM
  Can you explain the following terms please:

4 * 3 in C(4, 1) × 4 × 3.

C(4,1) in C(4, 1) × 2 × C(4, 1)

C(3, 1) × 2 in C(4, 3) × C(4, 2) × C(3, 1) × 2

C(2, 1) × C(3, 1) in C(4, 3) × C(4, 2) × C(2, 1) × C(3, 1) × 2