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Permutation & Combination !!!!!! help
by Saurabh Koshti - Saturday, 26 February 2011, 11:25 AM
  Q)  How many 6-digit numbers contain exactly 4 different digits ?

i) 4536
ii) 2,94,840
iii) 1,91,520
iv) none

How to solve this question ???
Re: Permutation & Combination !!!!!! help
by TG Team - Saturday, 26 February 2011, 02:03 PM
 

Hi Saurabh smile

Let the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.

So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).

Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10*9*8*7*5*13 = 327600 .

But is this the final answer? .....certainly not smile

See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.

Thus the answer will be 9*9*8*7*5*13 = 294840. smile

Kamal Lohia     

Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Saturday, 26 February 2011, 02:41 PM
  Thank you so much!!!!!!smilesmile
Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Friday, 4 March 2011, 12:55 PM
  another problem !!!!!

Q)  In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
Re: Permutation & Combination !!!!!! help
by TG Team - Friday, 4 March 2011, 03:41 PM
 
Hi Saurabhsmile

It is simply: 103C3 - 352C3 = 110551. smile

Amazed. Now see the explanation part.
Out of total 250 marks, you want to score 150 marks. So number of ways of scoring 150 is equivalent to not scoring the remaining 100 marks.
So let's say we have 4 boxes a, b, c, d such that capacity of a, b, c is 50 each and that of d is 100. And all 4 boxes are filled to its capacity and we are to remove 100 balls out of it. Does it sound better? smile
i.e. we are trying to find the whole number solution of: a + b + c + d = 100 with the restriction that none of a, b, c can be more than 50.
OK. Now without restriction, number of whole number solution of above equation is given by 103C3.
We need to subtract the cases when any one of a, b, c is more than 50 (i.e. 51 + x say, where x is a whole number). Remember no more than one of a, b, c can be greater than 50. Also I have taken it as 51 + x because that's how I ensure that one of a, b, c is certainly more than 50. And which one of a, b, c can be determined in 3 ways. Hence the final upfront expression.
Hope things are clear now. smile

Kamal Lohia  
Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Friday, 4 March 2011, 07:33 PM
  Thank you so much !!! again

 I  really appreciate this !!
Re: Permutation & Combination !!!!!! help
by anila . - Saturday, 12 March 2011, 05:49 PM
 

hello TG sir

sir i understood the 103C3  bt how we got 3 * 52C3...can you plz expalin ....why 52C3..plzzz can u explain more

and what if i want to do it other way a+b+c+d=150..then i will have 153C3 options and a,b,c should not be mor den 50 and d sholud not b more dn 100..then what exactly should i subtract

thank u

and saurabh gud questions..where can i get more such probability  and P&C questions.I am weak in probability and P&C section so i might help me..thank u

Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Thursday, 17 March 2011, 11:22 AM
  kamal sir can u please explain how to solve these type of question.... don't solve the question i will do that myself , jus give me som hint......thanks in advance!!

Q) The total number of integral solutions for (x,y,z) such that xyz = 24 ,is

i) 36 ii) 90 iii) 120 iv) none

Q) let A be the set of 4-digit number "a1a2a3a4"  where a1>a2>a3>a4
then how many values of A are possible ?
 
i) 126 ii)84  iii)210 iv) none
Re: Permutation & Combination !!!!!! help
by TG Team - Thursday, 17 March 2011, 01:48 PM
 
Hi Saurabhsmile

1) You are to just write 24 as a product of three integers (i - all three positive, ii - any two(i.e. selected in 3C2 = 3 ways) negative). So if I count the number of ways to write 24 as product of three natural numbers, that needs to be multiplied with 4 to get the final answer.
Now hint for that part: 24 = 23 * 3. That means you are to distribute powers of 2 and 3 in three numbers and then multiply them to get the ordered triplets.

2) We are to select 4 different digits out of available 10 and see that they can be arranged only in one possible way satisfying the given condition.

Hope it helps smile

Kamal Lohia
Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Friday, 18 March 2011, 09:21 AM
  thanks for the hint sir  !!!! here r my ans

1)  i m getting 30 pair  whose product is equal to 24 ; Now to get final ans i have to multiply it with 4 i.e 30 x 4 = 120 .

2) we can select 4 diff digits out 10 digit in 10C4 ways and then these four digit can b arranged in  only one way so ans is .. 10C4 X 1 = 210


***Correct me if i m wrong***
Re: Permutation & Combination !!!!!! help
by TG Team - Friday, 18 March 2011, 10:12 AM
  Good work Himanshu smile

Both answers are correct. smile

Kamal Lohia
Re: Permutation & Combination !!!!!! help
by Saurabh Koshti - Friday, 18 March 2011, 10:42 PM
 

A conference attended by 200 delegates is held in a hall. The hall has 7 doors,marked A,B,......,G. At each door,an entry book is kept and the delegates entering through that door sign in it in the order in which they enter . If each delegate is free to enter any time and through any door he likes, how many different sets of seven lists would arise in all ? (Assume that every person signs only at his first entry)

 

 

i) 206C6 ii) 199P5 iii) 199C5 iv) 206P6 (ans 206P6)

kamal sir what is the ans to this question ?? i m getting p(206,200)

not p(206,6)


Re: Permutation & Combination !!!!!! help
by TG Team - Thursday, 5 May 2011, 01:30 PM
  Hi Saurabh smile

Yes it will be P(206, 200) = 206!/6!

Kamal Lohia
Re: Permutation & Combination !!!!!! help
by nikita dhanuka - Wednesday, 11 May 2011, 11:17 PM
  Hi Kamal smile

great explanation! just one doubt...can we also approach the sum like this?
a + b + c + d = 150
with the restriction that a, b, c cannot be more than 50 and d cannot be more than 100.
therefore, the total no. of ways =
C(153,3) - 3*C(102,3) - C(52,3) = 48076 clown

i have understood your approach, but why is this not working sad sad sad
Re: Permutation & Combination !!!!!! help
by TG Team - Friday, 13 May 2011, 03:36 PM
  Hi Nikita smile

Please elaborate your expression so that it is easier to comment on.

Kamal Lohia
Re: Permutation & Combination !!!!!! help
by nikita dhanuka - Wednesday, 25 May 2011, 05:11 PM
  Hi Kamal smile

here is the question again..
Q) In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?

Soln: 60% marks in aggregate means a total of 150.
i have assumed a, b, c the marks in the first 3 papers and d as the marks in the fourth paper.
so, a+b+c+d = 150
under the restriction that a(max)=b(max)=c(max)=50 & d(max)=100.
then, i figured out the total no. of ways as

C(153,3) - 3*C(102,3) - C(52,3) = 48076

C(153,3) is the total no. of ways of distributing 150 amongst a,b,c,d without any restrictions.now, i am looking to subtract the cases where any of a,b,c is more than 50, or d is more than 100.
Assuming that a=51, we can distribute 99 among the 4 groups in C(102,3) ways. i have multiplied C(102,3) with 3 to take care of similar cases for b and c.
Lastly, assuming d=101, we can distribute 49 among 4 groups in C(52,3) ways. these cases are to be subtracted as well.

this is how i figured out the solution. i have understood your explanation. you approached the sum by thinking about the 100 marks which have not been scored. now, when i am applying similar logic and thinking about the 150 marks which have been scored, then why are the answers not matching up?? dead
Re: Permutation & Combination !!!!!! help
by Rajasekaran Rajaram - Friday, 27 May 2011, 08:08 PM
  kamal,

Thus the answer will be 9*9*8*7*5*13 = 294840.

I was thinking 9*9*8*7*4*5 to be the possibility since the question is asked for exactly four digits. So the last two digits are to be selected from the previously selected digits only.

Please point out the error in my logic.

Please forgive me if i sound too silly sad
Re: Permutation & Combination !!!!!! help
by saurabh katiyar - Sunday, 5 June 2011, 05:37 PM
  Hi Kamal,
A query was generated in my mind for the very first question of this topic raised by Saurabh Koshti.
you have selected 4 digits but it should be 10C6 because you r making a six digit no and than all those digits, u r arranging in 6! ways and dividing by 2!*2! because two nos. are repeating or It can be 10P6/2!2!.
If I am wrong, please help me to correct
Re: Permutation & Combination !!!!!! help
by ankita bhattacharya - Sunday, 28 April 2013, 01:38 PM
  can you pls explain me how is it 103c3..what is the idea behind it?
Re: Permutation & Combination !!!!!! help
by sushmita nigam - Saturday, 10 August 2013, 12:28 PM
  why 9/10 of the original answer? plz explain
Re: Permutation & Combination !!!!!! help
by Deepankar Manduri - Monday, 9 September 2013, 12:54 PM
  hello sir,
i am preparing for e-litmus exam
in this above problem i am not getting how c(4,2) and c(4,1)came .
can you elaborate this thing through an easy example... mixed
Re: Permutation & Combination !!!!!! help
by sneha singh - Monday, 8 September 2014, 01:19 AM
  Sir how u got dat c(103,3)
I didn't get it..can you please explain
Re: Permutation & Combination !!!!!! help
by Kundan Pandey - Thursday, 15 October 2015, 10:58 AM
  how 103c3 came and 52c3?? pls explain , am unable to get it
Re: Permutation & Combination !!!!!! help
by saquib saquib - Monday, 19 October 2015, 03:30 AM
  Using the famous rule for distributing n identical things to r different person/grp: n+r-1Cr-1
Here we have to distribute 100 marks to 4 entity therefor 100+4-1C4-1=103C3