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Permu n Combi - I
by vamsi krishna - Wednesday, 3 September 2008, 07:45 PM
 

Frns,

Lets solve all the Permutation n Combinations problems given by TG in Lessons Section - 10 a Day... What Say !

VaMsI

here are the first Ten :


Re: Permu n Combi - I
by Floydian _iva - Wednesday, 3 September 2008, 11:05 PM
  1. 5C0+ 5C1+5C2+5C3+5C4+5C5 = 2^5 = 32 (A)
    (Sets that contains one element: 1 (i.e. singleton with only one element '6')
     Sets that contains two element: 5C1 (Fix one as '6', choose one more from       rest 5 (exclude 7)...
     Similarly do it for sets that contains 3,4,5 6 elements)
2. The number would look like ABCCBA
    A can't be zero, cause if it is then it will be no longer a six digit number (in 
    common sensical terms)
    So answer would be: 9*10*10 = 900 (b)
3. I got the hold of it. The answer is 0 (b).
    It's simple. The addition is 2+5+7+8+8 = 30. It has to be divisible by 3.
4. 8C1+8C2 = 8+28 = 36 (a)
    Will do the rest tomorrow. The post was not looking good with no reply.
    Getting sleepy smile It's okay!  
    Good night!





  
 
                     
Re: Permu n Combi - I
by Tuhin Banerjee - Thursday, 4 September 2008, 05:41 PM
 

 Answers are :

1. 32               2.  900         3. 60        4. *****

5. 17999         6.   36          7. 3         8. 4

9. 2              10. 256  

kindly let me know the correct answers .

 Thanks

Tuhin

 

 

Re: Permu n Combi - I
by vamsi krishna - Thursday, 4 September 2008, 06:43 PM
 
Tuhin

4. Am gettin 26^3 . 220
7. donno....

Re: Permu n Combi - I
by Floydian _iva - Thursday, 4 September 2008, 09:35 PM
  Sorry the last one was the answer of question number 6.
Re: Permu n Combi - I
by Floydian _iva - Thursday, 4 September 2008, 09:47 PM
  The answer of q.4 is 26^3*220 only.
I am getting the same answer.
Re: Permu n Combi - I
by vamsi krishna - Saturday, 6 September 2008, 11:36 AM
  Hey
Can Sumone Throw sum light on PROB 4 n 7?
Re: Permu n Combi - I
by bimal mohan - Sunday, 7 September 2008, 05:15 PM
 

Prob 4:   I  think  the  statement  of   the   question  needs  slight   change  ....it   should   be  "each  digit   must   be   greater   than the  one  preceding it  "   ,   then only   we   can   zero   in  on  third    option (c) 26^3*120

Prob 7:  if   we  allocate  each  color  ,   we  can  see,  two  colors   must  be   used   twice  ( opposite  side ) ,   so  3C2  =3   ways

                  2R ---2B----G  , 2R---2G----B,2G----2B----R

TG  sir  ,  pls  confirm  the   answers.

Bimal

Re: Permu n Combi - I
by vamsi krishna - Sunday, 7 September 2008, 05:26 PM
  bIMAL.
regarding Prb-4
if TG sir's intension was "each digit must be greater than the one preceding it" than option (c) 26^3*120 snugly fits the question.

but
Can u calculate for given case (each digit must be equal to or greater than the one preceding it ) ?

Prob-7
Qustion is abt different ways of painting colours...
but u hav only considered selection of colors leaving aside distribution...!
Pl clarify

VaMsI
Re: Permu n Combi - I
by bimal mohan - Sunday, 7 September 2008, 05:57 PM
 

Q4  :   In  the   given  case  ,   ans   should   be   26^3*220....here   i'm   getting  a  generalised   form  of summation[ (n+1)*n/2] to   get  220.

Q7: As  it's  a  regular   pentagon  ,  so  if  we   allocate  colors in  the  particular   pattern  ,  picture   remains  same  .

pls  suggest  if   i'm  wrong.

Bimal

Re: Permu n Combi - I
by vamsi krishna - Sunday, 7 September 2008, 06:24 PM
  @Bimal

Q7.Allocation of colors to sides results in more than 3
cases. I am giving 4 cases to illustrate da same.

BRBGR
RGRBG
BRGBR
GRGRB (list goes on)

TG Sir pl come into picture!
Re: Permu n Combi - I
by bimal mohan - Sunday, 7 September 2008, 06:36 PM
 

hi   vamsi  , 

   BRBGR <----->BRGBR    & RGRBG<---->GRGRB     become   identical....pls   put   them  on  the   fig  of   regular   pentagon.

 



 

Re: Permu n Combi - I
by vamsi krishna - Sunday, 7 September 2008, 06:52 PM
  itz Hard 2 digest

Thanks Bimal...
Re: Permu n Combi - I
by vamsi krishna - Tuesday, 9 September 2008, 02:26 PM
 

Consoliated Answers:

1. 32

2. 900

3. 60

4. 26^3 * 220

5. 17999

6. 36

7. 3

8. 4

9. 2

10 256

VaMsI

Re: Permu n Combi - I
by vamsi krishna - Tuesday, 9 September 2008, 02:43 PM
 

Here goes the next ten,

Post ur ans only...

we'll compare our answers and discuss da insane ones

VaMsI


Re: Permu n Combi - I
by CATendra 2008 - Tuesday, 9 September 2008, 03:43 PM
  Hi Vams,
I am getting the  following answers.
11)2240
12)50
13)I couldn't solve this question.No clues!!!evil
14)I
15)35
16)20
17)240
18)9
19)Help needed!!
20)315

Lets discuss the approach for the 13th question..
10C4 -(No. of cases where 2 are together + when 3 are together + when 4 are together)
2 are together : 9 * 8C2 (here there can also be cases where 3,4 and 5 are together)
But 10C4 < 9 * 8C2angry.
There is some mistake i am committing which i am unable to catch hold of.
Re: Permu n Combi - I
by vamsi krishna - Tuesday, 9 September 2008, 05:22 PM
  Catender,

Q19)
In the expansion of (a1 + a2 + a3 +.......... + ar)^n

there are C(n+r-1,r-1) terms

Try for small values of r and n

VaMsI
Re: Permu n Combi - I
by CATendra 2008 - Tuesday, 9 September 2008, 05:42 PM
  Thanks Vamsi,
u took me to the good old days of IIT preparation..smile
Re: Permu n Combi - I
by vamsi krishna - Wednesday, 10 September 2008, 12:11 PM
  Catendra

18) i wenton physically counting , is there any other method?
("required ways = Total ways w/o condition -cases that donot satisfy condition " is also getting lengthier)

VaMsI
Re: Permu n Combi - I
by CATendra 2008 - Thursday, 11 September 2008, 04:50 PM
  Hi Vamsi,
if u fix any number , then the remaining 3 can be arranged in 3! ways out of which exactly half will satisfy the condition.Hence,
3*3!/2 = 9
I think you can also generalize this for bigger numbers for eg , for a  5 digit number if there is a similar limitation then the number of ways will be
4*4!/2 = 48.
Also what should be the approach for Q13?
Re: Permu n Combi - I
by vamsi krishna - Thursday, 11 September 2008, 05:45 PM
  But catend,

I verified for 5 digit numbers (Took first digit wrong and jotted down all 24 combinations and crossed those which match with original digit)


there are only 44 cases in which u can arrange all digits so that no digit goes to right place.

Seemz a program needs to written to check for higher numberz !

VaMsI
Re: Permu n Combi - I
by Floydian _iva - Thursday, 11 September 2008, 06:12 PM
  19: (a) 136.
Solution: See, any term will look like: a^x*b^y*c^z
such that x+y+z= 15
and x is an integer belongs to the set [0,15]. The same goes with y and z.
So the answer will be coeff of x^15 in (x^0+x^1+x^2+...+x^15)^3
                                             = C(15+3-1, 15)
                                             = C(17,15)
                                             = 136

Avi
Re: Permu n Combi - I
by Bishweshwar Pradhan - Friday, 12 September 2008, 08:32 AM
 

Could not get the steps for 10th one. Please post it.

Thanks

Bishweshwar

Re: Permu n Combi - I
by bimal mohan - Friday, 12 September 2008, 08:56 AM
 

hi  all,

Q18:   let  us   consider  that  the   digits  2,3,1,4   are   destined   to  thousand,hundred,tens   and   unit   places   respectively   ......so   if   we   have   to  assign   that   all   of   them  placed   wrong  ------4![1-1/1!+1/2!-1/3!+1/4!] = 9ways

Q13:   leaving   aside   4   persons,   remaining   =6    so   we   can   place   this   4   guys   at   (6+1)= 7   places   ----7C4 = 35  ways.

pls   suggest   if   i'm  wrong!

Bimal

Re: Permu n Combi - I
by sunil garg - Friday, 12 September 2008, 01:15 PM
 

vamsi/ anybody

please explain the question number 4 !!!!!!!

its getting OVER HEAD TRANSMISSION

Regards

Sunil Garg

Re: Permu n Combi - I
by vamsi krishna - Friday, 12 September 2008, 01:49 PM
  Hi Sunil,

4 or 14?

VaMsI
Re: Permu n Combi - I
by sunil garg - Friday, 12 September 2008, 03:00 PM
 

PLEASE PROVIDE SOME HINTS OR APPROACH TO ALL  QUESTIONS AS WELL

Re: Permu n Combi - I
by CATendra 2008 - Friday, 12 September 2008, 03:13 PM
  @VaMsI
you've got it right.44 is correct.Look at the soln provided by Bimal.It's simple derangement formula.(Guess shud have read thru the entire article...grin)

@bimal
Thanks Bimal for the derangement  answer.

For question 13 , i have a query
If we change the question as how many ways can you select 4 out of 10 people such that there are atleast 2 persons between every 2 selected.
can we use a similar approach for the above problem..?(A tricky one i guess!!)
Re: Permu n Combi - I
by vamsi krishna - Friday, 12 September 2008, 04:27 PM
  Thanks Bimal n Catendra

I completed overlooked this Derangement formula..

I further wenton to check meaning of Disarrangement n Derangement

Disarrangement : A condition in which an orderly system has been disrupted

Derangement : A state of mental disturbance and disorientation

The former suits better i suppose...TG Sir!

VaMsI
Re: Permu n Combi - I
by Tuhin Banerjee - Friday, 12 September 2008, 05:00 PM
 

11) 2240   12)50   13)35    15)35   

16)12      17) 168    18) 9    19)136   20)315

 

Re: Permu n Combi - I
by vamsi krishna - Friday, 12 September 2008, 05:58 PM
  HEY Tuhin

watz ur approach for 13th

VaMsI
Re: Permu n Combi - I
by sunil garg - Saturday, 13 September 2008, 09:44 AM
 

hi vamsi

yaar please explain the qn number 4 and pentagon wala qn

regards

sunil garg

Re: Permu n Combi - I
by sunil garg - Saturday, 13 September 2008, 10:00 AM
  ANSWER OF QN 14 IS ............'S '......For Sunil
Re: Permu n Combi - I
by vamsi krishna - Saturday, 13 September 2008, 10:32 AM
 

Sory Sunil I was late in replyin

 

Q4)

first three positions are characters out of a-z, so 26^3 ways

next three are digits:

we need to form a 3 digit number out of 10 digits ,so that

a) each digit mst be greater than the preceeding number :
select 3 digits out of ten in C(10,3) and there is only one way of arranging them so sol: 26^3 . 120

b) each digit mst be greater than or equal to the preceeding number :
let the first digit is 0 then  the 2nd and 3rd digits can be
(0,0) TO (0,9) 10 cases
(1,1) TO (1,9)  9 cases
(2,2) TO (2,9)  8 cases
.
.
(8,8) TO (8,9)  2 cases
(9,9)               1 cases

if first digit is  0, then 2nd and 3rd give 10+9+8+7+.....2+1 cases

Similiarly if first digit is  1, then 2nd and 3rd give 9+8+7+...1 cases

....

...

..

.

Similiarly if first digit is  9, then 2nd and 3rd combine to 1 case

i,e Sigma(n(n+1)/2), n = 1 to 10
expand and apply u ll get 220

Hence  26^3 . 220

TG sir pl correct me if i am wrong

Q7) was well explained by Bimal Checkout previous posts...and try your part in interpreting it ..it took sometime for me to grasp...

VaMsI

Re: Permu n Combi - I
by vamsi krishna - Saturday, 13 September 2008, 12:43 PM
  Guyz,

14) 'T'

VaMsI
Re: Permu n Combi - I
by sunil garg - Saturday, 13 September 2008, 02:04 PM
 

thanx vamsi

for question number 4!!!!smile

what about pentagon question

regards'

sunil

Re: Permu n Combi - I
by vamsi krishna - Saturday, 13 September 2008, 02:14 PM
  @ bishweshwar

10) there are 2 ways to move around each circle

8 adjacent circles implies 2^8 = 256 ways

VaMsI
Re: Permu n Combi - I
by Floydian _iva - Monday, 15 September 2008, 02:09 PM
  Next ten Sir?
Regards
Avi
Re: Permu n Combi - I
by sunil garg - Monday, 15 September 2008, 05:43 PM
 

hi vamsi

please tell the approach for solving the question number 15 and 18 ( if possible almost for all qns)

what is the source of these qns these qns are really mind opening

thanx and regards

sunil garg

Re: Permu n Combi - I
by Rishi Kapoor - Tuesday, 16 September 2008, 01:35 AM
  11- A
12-B
13-B
14-B
15-A
16-B
17-C
18-D
19-Don't know
20-C

...RK...
Re: Permu n Combi - I
by ambar patil - Tuesday, 16 September 2008, 07:14 PM
 

Vamsi / anyone  ,

Can you please throw some light on these questions before moving to 3rd set ?

5 .( from prev set ). there are 5 digits . 1 3 5 7 9 . keeping 1 as the first digit We get 4*4*4*4 i.e 256 numbers that start with 1 . So how how can the 500th number start with 1 . Or is it the 50th number that is being asked ?

12 .

15 . Could not get ' If we are intrested only in the country they are representing ' .

16 . got 10 ways in which A could win irrespective of who wins the first match ( A - 6 cases , B - 4 cases ) . So total ways 20 . But is brute force the correct way to go for it ?

20 .  

Re: Permu n Combi - I
by ambar patil - Tuesday, 16 September 2008, 07:17 PM
 

Vamsi / anyone  ,

Can you please throw some light on these questions before moving to 3rd set ?

5 .( from prev set ). there are 5 digits . 1 3 5 7 9 . keeping 1 as the first digit We get 4*4*4*4 i.e 256 numbers that start with 1 . So how how can the 500th number start with 1 . Or is it the 50th number that is being asked ?

12 .

15 . Could not get ' If we are intrested only in the country they are representing ' .

16 . got 10 ways in which A could win irrespective of who wins the first match ( A - 6 cases , B - 4 cases ) . So total ways 20 . But is brute force the correct way to go for it ?

20 .  

Re: Permu n Combi - I
by ambar patil - Tuesday, 16 September 2008, 09:13 PM
 

arrghh !! i got my error for the 5th one , it should be 5*5*5*5 and not 4 . So i get 500th digit as 17999 .

11 _ _ _ .  now these are 125 nos. similarly 13 _ _ _  , 15 _ _ _ , 17 _ _ _ . So total 125 * 4 = 500 and the 500th ll be 17999 .

 

Re: Permu n Combi - I
by Rishi Kapoor - Tuesday, 16 September 2008, 09:55 PM
  Question-
In how many ways can one or more of 5 letters be posted into 4 email boxes, if any letter can be posted into any of the boxes?
1-54
2-45
3-55 -1
4-45 -1

...RK...
Re: Permu n Combi - I
by ambar patil - Tuesday, 16 September 2008, 10:27 PM
 

each letter can be handled in 5 ways . Either post it in one of the 4 email boexs or do not post it .

So ans option 4 -  4^5-1

Re: Permu n Combi - I
by Rishi Kapoor - Wednesday, 17 September 2008, 12:56 PM
  Ambar,
In that case the answer should be 55-1....and that's the answer indeed!
Thanks
...RK...
Re: Permu n Combi - I
by ambar patil - Wednesday, 17 September 2008, 01:45 PM
  lolz ... yes !!
Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 02:49 PM
  AmBeR


12)
Last digit can be
1-so remaining 2 digits be selected out of 8 in 8C2 and arrange in only one way
3-so remaining 2 digits be selected out of 6 in 6C2 and arrange in only one way
5-so remaining 2 digits be selected out of 4 in 4C2 and arrange in only one way
7-so remaining 2 digits be selected out of 2 in 2C2 and arrange in only one way
9- No chance
Sumup to 50

15)
Try this question AmBeR
In how many ways can u arrange 4 apples and 3 oranges?
samething was asked in Q15

16)
cases for each player
3 matches -> hav to win three successively..only one way
4 matches -> _ _ _ W , the blanks can be filled with all permutations of W,W,L
in 3!/2! ways
5 matches->_ _ _ _ W , the blanks can be filled with all permutations of W,W,L,L
in 4!/(2!.2!) ways

so tatal 10 ways for each player
for two players 20 ways

20)
select 3 letters out of 7 letters in C(7,3) and send them to their correct envelopes in 1 way
Then u r left with 4 letters, 4 envelopes
now question states tat there are 9 permutations for 4 letters in 4 envelopes
hence ans: 7C3.1.9

Pl commentz if i am wrong !

VaMsI
Re: Permu n Combi - I
by ambar patil - Wednesday, 17 September 2008, 03:22 PM
 

Nice explanation , Bro !! gaat it . smile

Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 03:27 PM
  Thanks Amber

Hey

any Takers for Q13 n Q14?

VaMsI
Re: Permu n Combi - I
by Amit wadhwa - Wednesday, 17 September 2008, 04:05 PM
 

hey Vamsi

how would you get 220 in Q4.

rgds/Amit

Re: Permu n Combi - I
by Floydian _iva - Wednesday, 17 September 2008, 04:52 PM
  14.
The word is: MATHEMATICS
There are 2 A's, 1C, 1E,1H,1I, 2M's, 1S, 2T's.
Now the first word will be AACEHIMMSTT (just simply I have put in alphabetical order).
Like the way words are arranged in dictionary, just consider
AACEHI  MMSTT, this MMSTT can be arranged in 5!/2!*2! = 30 ways.
So with the same AACEHI we can form 30 words...
and the 30th word will look like AACEHI TTSMM (I hope it is clear)
Since we have to find the 50th word I am taking AACEHI out...because if I take AACEH out then with this AACEH we can form 6!/2!*2! = 180 words.
Now, now the 30th word would be AACEH I TTSMM.
Replace I by M and after arrange the rest in alphabetical order to get the next word i.e. AACEH M IMSTT. (31st)
Now look at AACEHMI MSTT...again with the first AACEHMI you can form 4!/2! = 12 words.
So 42nd word will be AACEHM I TTSM.
replace I by M to get AACEHM M ISTT as 43rd word.
Take AACEHMMI out ... and on the same fashion AACEHMM I TTS will be the 45th word.
Replace I by S, to get AACEHMM S ITT as 46th word.
48th word: AACEHMMSTTI
49th word: AACEHMMSTIT
50th word: AACEHMMSITT
so the tenth word will be T.

Regards Avi.








Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 05:09 PM
  Yep buddy,
I know it will be u who will be addressing Q14..
I 2 got da same ans
I substituted numbers for characters after finding 42nd word and raced on.

but still i felt it wasnt a 2 minute problem.

VaMsI
Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 05:03 PM
  Hi Amitji

I hav already posted my solution.
jst scroll down furthr.

VaMsI
Re: Permu n Combi - I
by Amit wadhwa - Wednesday, 17 September 2008, 05:06 PM
 

hey rishi

how come 55-1 .. it should be 54-1...

 

Amit

Re: Permu n Combi - I
by Floydian _iva - Wednesday, 17 September 2008, 05:07 PM
  Of course...man...I posted it here because atleast people should be aware of this type of problem. One good suggestion is that never attempt such type of problems in CAT. It's not JEE (even JEE is no longer same now!).
All you have to do is to select the easiest problems and crack it in 45 minutes. Our target is to score more, we need not show our attitude that we can solve these type of problems. They are looking for managers not puzzle solvers.
in good faith
Avi
Re: Permu n Combi - I
by Amit wadhwa - Wednesday, 17 September 2008, 05:12 PM
 

Thnx vamsi..

i got the solution for q4.

could you pls tell me how to solve Q6.

Amit

Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 05:47 PM
  amit

Same reply again

jst scroll down...

Check da solution given for Q19

VaMsI
Re: Permu n Combi - I
by vamsi krishna - Wednesday, 17 September 2008, 05:53 PM
  hmm..Sunil..

15)Jst scroll down sumwhere i hav discussed.
18) disarangement formula (Thanks 2 TG sir)

Source: Wat else...TG site
at da end of this lesson
http://totalgadha.com/mod/forum/discuss.php?d=3537

VaMsI
Re: Permu n Combi - I
by Floydian _iva - Wednesday, 17 September 2008, 11:54 PM
  13.
 x1 |  x2 |  x3  |  x4 | x5
Pipes indicate the 4 persons...and x1, x2, x3, x4 and x5 are the number of candidates in between.
Now,  x1 can be 0, 1, 2, 3
         x2 can be 1, 2, 3, 4
         x3 can be 1, 2, 3, 4
         x4 can be 1, 2, 3, 4
         x5 can be 0, 1, 2, 3
with the condition x1+x2+x3+x4+x5 = 6
So the answer will be coeff of x^6 in (1+x+x^2+x^3)^2*(x+x^2+x^3+x^4)^3
                                           = coeff of x^3 in (1-x)^(-5)
                                           = C(5+3-1,3)
                                           = 35
Regards
Avi

Re: Permu n Combi - I
by CATendra 2008 - Thursday, 18 September 2008, 06:01 PM
  Q13
Take the four people out.We are left with 6 people and 7 places to fit in those 4 asses.This can be done in 7C4 ways = 35

Re: Permu n Combi - I
by vamsi krishna - Friday, 19 September 2008, 12:50 PM
  @ avi
Again am lost

@ Catendra
Hey Gr8 solution Bro

Thks

VaMsI
Re: Permu n Combi - I
by vamsi krishna - Friday, 19 September 2008, 01:07 PM
  Consolidated Ans for SET-II

11.2240

12.50

13.35

14.T

15.35

16.20

17.240

18.9

19.136

20.315


VaMsI
Re: Permu n Combi - I
by vamsi krishna - Friday, 19 September 2008, 01:15 PM
 
N da next Set is here...

Am sure "avi" is eagerly lurking around for this set

VaMsI
Re: Permu n Combi - I
by Floydian _iva - Friday, 19 September 2008, 02:14 PM
  1. 204
2. 32
3. 35,120, 10(4!)
4. 26!/2
5. 45, 21, 35
will do the rest later.
Regards
Avi
Re: Permu n Combi - I
by ambar patil - Friday, 19 September 2008, 02:32 PM
 

21. 204

22. 32

23. a. 7!/3! = 840

     b. 5!

     c. 4!

24. 26!/2

25. a.10C8 = 10C2 = 45

      b. 7C5 = 7C2 = 21

      c.We will consider two cases . 4 out of 1st 5 are answered & 1st 5 are   answered . 

        So , 5C4 X 6C4 + 5C3 = 85 .

 

Re: Permu n Combi - I
by Rishi Kapoor - Friday, 19 September 2008, 03:53 PM
 

26- 45

27-11804750000 ????

...RK...

Re: Permu n Combi - I
by Floydian _iva - Friday, 19 September 2008, 04:45 PM
  Ambar,
23 rd first part I got wrong...you are correct ...its 840 only.
Regarding 25(c):
in the first case, when you are choosing 4 out of first 5, and 4 out of the remaining 5, number of ways will be...5C4 x 5C4...what you are doing is that you are counting it twice. 5C4 X 6C4 includes the second case too. Got me??
Regards
Avi

      
Re: Permu n Combi - I
by ambar patil - Friday, 19 September 2008, 04:52 PM
  ah ... i see . what a trap ! smile  
Re: Permu n Combi - I
by Floydian _iva - Friday, 19 September 2008, 05:23 PM
  Rishi
Q 26. I am getting the answer (8!/5!)
My approach: Assume the block to be one single object (all are together).
Now there are 8 objects, 3 blocks and 5 empty seats.
All the empty seats looks alike, so the number of ways of arranging them will be
8!/5!.
Its like B1, B2, B3, E, E, E, E, E. What say?
Regards
Avi
Re: Permu n Combi - I
by Tuhin Banerjee - Friday, 19 September 2008, 07:00 PM
 

Hi Vamsi,

Sorry being late, approach for the 13th no. same as Bimal mentioned.

Re: Permu n Combi - I
by bimal mohan - Friday, 19 September 2008, 09:16 PM
 

Very  well   explained   ....but  i   think     " from 48th word: AACEHMMSTTI    to  49th word: AACEHMMSTIT"   ,   there  is   something  wrong     bcoz  AACEHMMSTIT   can   not   come   after   AACEHMMSTTI  .....

if  we   arrange   the  letters  :   AACEHIMMSTT .......  as  we   are   bothered   about  50th   word  ,  we  need  to  pay   attention   to  the   arrangement   of   last  6  letters ...                                                                               

      IMMSTT ----------------->ITTSMM [  5!/2!*2! =30 words] -------->MIMSTT( next word)-------->MITTSM [ 4!/2!=12]----------->MMISTT( next word)-------->MMITTS[3!/2!=3 ]-------->MMSITT( next word)-------->MMSTTI[3!/2!=3 ]-------->MMTIST ( 49th word) --------->MMTITS( 50 th   word)

 10 th Letter : T

pls   suggest  if  i'm   wrong.

Bimal

Re: Permu n Combi - I
by bimal mohan - Saturday, 20 September 2008, 12:22 AM
 

21)204         22)32       23)840/120      24)26!/2         25)45/21/35        26)56        27)11804750000      30)19

Bimal

 

Re: Permu n Combi - I
by Rishi Kapoor - Saturday, 20 September 2008, 01:27 AM
  IVA,

I was wrong.......so were u...I suppose!
Right answer should be (8!)/ (5! * 3!) coz just like u arranged 5 empty seats, we also have to arrange 3 blocks. They also look same.
Well thanks, I commit the blunder by not arranging anything at all.
...RK...
Re: Permu n Combi - I
by Floydian _iva - Saturday, 20 September 2008, 07:54 PM
   Bimal...you are absolutely right!
Just after posting, I realized. But by that time it was too late to edit.
Thanks for the correction. I hope, I will not commit same error on D-day.
smile
God bless you.
Avi.


Re: Permu n Combi - I
by nitin taluja - Friday, 26 September 2008, 03:13 PM
  Hi Vamsi,

4) 120 * 26^3 would be the answer

26^3(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36) use the foormula n(n+1)/2 to the get the result
for example in case of 9 only one number can be formed which 999 , in case of 8 at max four(because we need to fill two places and we have two numbers 8,9) numbers can be formed out of which 3 numbers are valid because of the condition similarly the others




Re: Permu n Combi - I
by vamsi krishna - Friday, 26 September 2008, 03:34 PM
  nitin

Scrolldown further, i hav mentioned my approach for Q4.

VaMsI
Re: Permu n Combi - I
by nitin taluja - Saturday, 27 September 2008, 09:42 AM
  Hi Vamsi,

I have gone through ur approach,its been good. The answer is surely (220*26^3)
if we consider that number can be started with zero which do in terms of licence plates but if we consider a digits can never start off with a zero then answer would reduce to 120*26^3 as the mentioned answer.

Hi ... TG please clarify on this questions ....

Regards
Nitin

Re: Permu n Combi - I
by VIKAS NIGAM - Saturday, 27 September 2008, 03:52 PM
 

Hi Vamsi,

My answers are:

21) (8x8) + (7x7) + (6x6) ..... (2x2) + (1x1) = 204

22) 4!(beginning with C) + 3!(beginning with FC) + 1(FOCSU) + 1(FOCUS) = 32

23) i) 7!/3! =  840

    ii) Taking 'UUU' as one alphabet, 5! = 120

   iii) _1_2_3_4_ Filling U's in the blanks in 5C3 ways and permuting N,S,A,L at positions    marked as 1,2,3 and 4 in 4! ways.So = 4!x10 = 240

24) b can be either on the left or on the right of e in any of the possible permutations. So half the times it will be on the left and half the time it will be on the right, therefore = 26!/2

25) i) 10C8 = 45

     ii) 1x7C5 = 21

     iii) (5x5C4) + (1x5C3) = 35

26) a+5+b+5+c+5+d = 20 => a+b+c+d = 5 Here a,b,c,d are non-negative integers. Number of ways = possible solutions to this eqn. = (5+3)C3 = 56

27) (25C4x46C3)+(25C3x25C1x47C1)+(25C2x25C3x48C3)+(25C4x50C3)

28) a) (2n-1)x(2n-3)x(2n-5)x.....3x1

29) 20! x 220

 

Do let me know the correct answers.

 

Vikas 

Re: Permu n Combi - I
by Rahul Mathur - Tuesday, 30 September 2008, 03:19 AM
 

Hi Vamsi.
   As for  Q.14, my approach is as below:

 First I kept the first 6 letters constant as AACEHI. So there will b a total of 30 words starting with AACEHI (5!/2!*2!=30)
 
 Next I interchanged the first M & I and kept 7 letters constant as AACEHMI. So, now there will b a total of 12 words starting with AACEHMI (4!/2!=12) and the 42nd word being AACEHMITTSM.
 
 Now, the 43rd word will be AACEHMMISTT.
 Going along the same way, 50th word comes out to be AACEHMMTITS. So, my word is different from avi's.
 Still, the tenth letter is T.

Re: Permu n Combi - I
by sunil garg - Wednesday, 1 October 2008, 04:36 PM
 

hi friends

the following pblm is looking simple but its giving me pain

please solve

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

answer choices:120,240,480.960

Re: Permu n Combi - I
by vamsi krishna - Wednesday, 1 October 2008, 04:50 PM
  480

VaMsI
Re: Permu n Combi - I
by Partha Ray - Wednesday, 8 October 2008, 03:32 AM
 

Hi TG/Vamsi,

Can you pls throw more light on the approach to this particular type of problem.

There is a guy named Gadhanand...who appeared an exam that consisted of 4 papers...the maximum marks of 3 of these papers is 50...while the max of 4th paper is 100... In how many ways can he secure 60% of marks in the exam in order to change his name from Gadhanand to Vidyanand???

Partha

Re: Permu n Combi - I
by vamsi krishna - Wednesday, 8 October 2008, 10:27 AM
 

Partha,

 

I Understood tat we need to find the number of ways in which Gadhanadh can score exactly 60 % :

 

a+b+c+d = 150 marks… ….(1) ( 60 % of 250 marks)

d=150-(a+b+c) where

0 <= a,b,c <= 50 …………..(2)

0 <= d <= 100……         ….(3)

 

Using (1)  (2) and (3) , 50 <= a+b+c <= 150

 

since a1+a2+a3+...+ar = n, where a1,a2,a3..are non-negative, has C(n+r-1,r-1) solutions, we have

a+b+c=  50 and (2) implies 52C2 solutions

a+b+c=  51 and (2) implies 53C2 solutions

.

.

.

.

a+b+c=  150 and (2) implies 152C2 solutions

 

total solutions = 52C2 + 53C2 + 54C2 + ......152C2

 

SOmeone correct me if I missed something

 

VaMsI

Re: Permu n Combi - I
by Partha Ray - Wednesday, 8 October 2008, 12:26 PM
 

Thanks Vamsi, the approach is perfect...

Cheers

Partha

Re: Permu n Combi - I
by ak _cat - Thursday, 9 October 2008, 03:15 PM
 

hi

pl  expalain explain q no-28,29&30 in detail

can somone explain the approach behind (cube & colour) questions.

Re: Permu n Combi - I
by Deepak Shukla - Friday, 10 October 2008, 12:01 AM
  could u plz explain how u get answr to question no. 05
Re: Permu n Combi - I
by Anshu Airan - Saturday, 11 October 2008, 06:30 PM
  Hey please provide explanation key for all the questions. sad
Re: Permu n Combi - I
by Anshu Airan - Saturday, 11 October 2008, 06:31 PM
 

someone please provide explanation key for all the questions

Re: Permu n Combi - I
by nirmesh sinha - Monday, 13 October 2008, 09:19 PM
 

Frnds ,can anybody help me in solving these 2 question

1. find the number of non-negative integral solution of |x|+|y|+|z|=15

2. find the number of non-negative integral solution of 2x+2y+z=10

Re: Permu n Combi - I
by vamsi krishna - Monday, 13 October 2008, 10:17 PM
  Nirmesh

1)136
2)21

Approach:
1)nothing but x+y+z = 15
C(n+r-1,r-1) = C(17,2)= 136

2)
2(x+y)+z = 10
z can take even values of 0,2,4,6,8,10 only for obvious reasons.
for z = 0, x+y = 5, C(5+2-1,2-1) = 6 solutions
for z = 2, x+y = 4, C(4+2-1,2-1) = 5 solutions
.
.
.
for z = 10 x+y = 0, C(0+2-1,2-1) = 1 Solutions
hence total solutions = 6+5+4+3+2+1 = 21


VaMsI
Re: Permu n Combi - I
by Vernika Singla - Tuesday, 14 October 2008, 11:12 AM
  While considering case b in the solution to the question 4 above, haven't we already taken greater than preceding no. in case (a)??
Re: Permu n Combi - I
by dhwani parikh - Wednesday, 5 November 2008, 07:43 PM
  i'm getting answer of 9th question 3... plz clarify it..
Re: Permu n Combi - I
by amit jain - Tuesday, 23 June 2009, 09:54 PM
 

Hi all,

 

Need the explanation for this question:

 

(a+2b+3c-5d+4e-f-2g-7h)^2 . Number of terms in this?

Re: Permu n Combi - I
by userdce . - Wednesday, 24 June 2009, 11:07 AM
 

@amit jain

while we expand any squared equation like urs then the terms would be sum of individual squares of each term and then product of two terms

like (a+b+c)2= a2+b2+c2+2ab+2bc+2ca

so general answer is

all individual terms + two terms out of all terms

=3+3C2

now with ur problem

answer= 8+8C2=8+28=36

let me know if wrong

 

there is beautiful explanation to ur question by DAGNY

http://totalgadha.com/mod/forum/discuss.php?d=199

Re: Permu n Combi - I
by aditya ranjan - Wednesday, 24 June 2009, 02:34 PM
  answer is right bt ur soln says AACEHMMSTIT to be before (49th)
AACEHMMSITT(50th) its wrong u can see it easily

last word (50th)according to me will be AACEHMMSTTI


correct me if m wrong smile big grin
Re: Permu n Combi - I
by satyarth pandey - Tuesday, 8 September 2009, 06:09 PM
  hi need the solution for q17....
i am doing some stupid mistake and need to know where am i going wrong
thanks
Re: Permu n Combi - I
by Deepika Agarwal - Thursday, 24 September 2009, 09:39 AM
  I'm getting it as 240 (5C2 * 4!)
Please let me know your approach.
Re: Permu n Combi - I
by satyarth pandey - Friday, 25 September 2009, 08:33 PM
  hi deepika,
 thanks for the reply....i'm getting the ans as 168...

my approach.....no. has to be 3 digit and less than 800 and must contain digit 5
hence 3 possible cases arise:

case 1:  _ _ 5 :    0,8,9 cant be the first digit hence 6 no.s can take first place and then 8 no.s can take the second place....hence 6*8= 48

case 2: _5_ :   similar to case 1...hence  6*8 = 48

case 3: 5 _ _ :   9 no.s can take the second place and 8 no.s the last hence
                         8*9 =72
summing up we get 168....

i missing something in my approach but cant make out...let me also know ur approach...may be i can find something in it

thanks & Regards
satyarth
Re: Permu n Combi - I
by Ashish Tripathi - Tuesday, 27 October 2009, 06:14 PM
 

Hi Vamsi,

Grt initiative dear. Well! recently, i started working on the sets posted  onthis link, tht is why asking about ques # 17 in set 2. I got 168 as the answer.

Sol: There are 3 cases of fixing 5:

a) 5 _ _ here we can fill 10s place in 9 ways and units place in 8 ways, so the total no. of ways = 9*8 = 72

b) _ 5 _ here hundreds place can be filled with 1 to 7 digits except 5, so no. of ways become 6, and units place can be filled with 0 to 9 digits except 5 and the hundreds digit, so no. of ways become 6*8 = 48 ways

c) _ _ 5 here also 6*8 = 48 ways,

so the total no. of ways = 72 + 48 + 48 = 168 ways.

Please suggest. 

Re: Permu n Combi - I
by KAMONASISH AAYUSH MAZUMDAR - Saturday, 31 October 2009, 04:51 PM
  whats the official answer/solutions for questions 1 through 20 ?

can anyone please PLEASE please write them down here? i would be very thankful.

please PM me on my TG id if someone does.

thanx in advance smile
Re: Permu n Combi - I
by bhanu nemani - Saturday, 6 August 2011, 04:57 PM
  Hi vamsi krishna,
this is bhanu...i'm about to give CAT this year...
your way of approach to the problems were really good..
thanks a lot smile smile smile
Re: Permu n Combi - I
by ketav sharma - Wednesday, 7 September 2011, 09:40 PM
  in 25(c) part
if we select 4 questions from first five then it is 5c4 = 5
and now as 6 questions are remaining from which we have to select 4 more then it is 6c4 = 15
15 * 5 = 75
How is it wrong ?
Re: Permu n Combi - I
by Pulkit Mehrotra - Friday, 14 December 2012, 10:47 PM
  can u please explain me the method to solve ques number 10...that is of 8 circles...
Re: Permu n Combi - I
by GAUTHAM thomas - Wednesday, 19 December 2012, 12:59 PM
  Hi all,

Can someone explain this plssss sad ?

In how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates?
TG sir had answered this and the answer was 28.
----------------------------------------------------------------
suppose if this question was changed such that, the first person receives at most 6, second person receives at most 7 and third person receives at most 8. Then this would be about
solving

A+B+C = 12 where A is less than or equal to 6
B is less than or equal to 7
C is less than or equal to 8

Now, what about this case? If the technique mentioned by TG is used, then we have to distribute 6 to A, 7 to B, 8 to C. So in total they have 6+7+8 = 21 chocolates. So we have to pick 9.
So this is about solving the equation A+B+C = 9. This is by the method of partition gives 11C2 ways or 55 ways.
But the solution mentioned says 45 ways

What is happening ? sad
Re: Permu n Combi - I
by ankit gaur - Wednesday, 19 December 2012, 07:25 PM
  Hi dear,

Given a + b + c = 12. As a, b, and c can take maximum value of 6. So, first i give all three six each and then remove a single 6 from all three.

So, solution for a + b + c = 12, where a,b,c can not be more than six will be the same as the solution for x + y + z = 6.

So, answer will be 8C2 or 28 ways.

But if the question is A+B+C = 12 where A is less than or equal to 6, B is less than or equal to 7, C is less than or equal to 8

Now, we have to distribute 6 to A, 7 to B, 8 to C. So in total they have 6+7+8 = 21. So we have to take 9 back.
So this is about solving the equation x + y + z = 9. So, by the method of partition gives 11C2 ways or 55 ways. But in this case method we can not take 9 or 8 or 7 back from x, same for y or z . So, this method is wrong or we can make it right by subtracting these cases.

Let do this in another way.

A + B + C = 12, total number of ways 14C2 or 91 ways.

Now as A can not be more than 6 than subtract all the cases in which A is more than 6. So, put A = 7+ x (A become more than 6) and subtract the cases for x + B + C = 5, number of ways 7C2 or 21 ways.

Same as B can not be more than 7 than subtract all the cases in which Bis more than 7. So, put B = 8+ x (B become more than 7) and subtract the cases for A + x + C = 4, number of ways 6C2 or 15 ways.

Same for doing C, A + B + x = 3, 5C2 or 10 ways.

So, answer will be 91- (21+15+10) = 45 ways.

Re: Permu n Combi - I
by GAUTHAM thomas - Tuesday, 25 December 2012, 03:47 PM
  Hi Ankit sir,

Thanks so much for the reply!! ... I've been looking for this explanation all over the place

I have a few doubts from what you said.. for the first question which TG had explained in the forum

1. In the question which TG had solved, A+B+C = 12 where A,B,C can take a maximum value of 6. This means, that A,B,C are all less than or equal to 6 right ?

2. Suppose if I changed the limits to -- > A,B,C are all less than or equal to 5. Even then the technique doesn't give the correct answer. The equations to be solved will be A+B+C=3

I assumed that this technique (mentioned by TG) could be used only if all 3 variables had the same upper limit. Am i wrong in thinking so ? (because in this we are removing a single 3 from each of A,B,C. The upper limits are 5. SO we can definitely remove a 3.. But it doesn't work) sad

3. Can you please explain how you arrive at the variables X,Y,Z ?

4. Can you throw more light on your statement - >
"But in this case method we can not take 9 or 8 or 7 back from x, same for y or z"

Thanks so much for the explanation. Pls help me in clearing this up
Re: Permu n Combi - I
by saurabh kumar - Friday, 16 August 2013, 12:14 AM
  n0 - 18     This is an dearrangment question 4!(1/2!-1/3!+1/4!) = 9

Re: Permu n Combi - I
by ankit SAHAY - Tuesday, 24 September 2013, 12:02 PM
  HEY, can some one pl;s explain Q.17?
the 3-digit number one...
thanx.
Re: Permu n Combi - I
by TG team - Thursday, 7 November 2013, 11:51 AM
 

Q.17

Considering 1st digit to be 5

2nd digit can be=0  then we have 8 options for third digit(1,2,3,4,6,7,8,9)

total options for 2nd digit=0,1,2,3,4,6,7,8,9=8

8*8=64

Considering  2st digit to be 5

1st digit =1

3rd digit can go from(0 2 3 4 6 7 8 9) = 8

now,first digit can go from 1,2,3,4,6,7(number should be less than 800)=6

total number possible=6*8=48

Considering  3st digit to be 5

1st digit =1 2nd digit =0 2 3 4 6 7 8 9 =8

1st digit can take 1 2 3 4 6 7(number should be less than 800) =6 values

total of 6*8=48

total=72+48+48=168

Re: Permu n Combi - I
by RASHMI BS - Friday, 31 January 2014, 06:51 PM
  can anybody explain me here

answer for the question 4 how it is 26^3 * 220??

my approach (26*26*26) for letters =26^3

and for digits how we need to calculate please explain

as i have thought it should be 9*8*7=504

and we need to subtract 504 from 729=so we get 225 .please correct me if am wrong
Re: Permu n Combi - I
by TG Team - Wednesday, 26 February 2014, 04:41 PM
 

Hi Rashmi smile

Here you need to form a number of three digits such that each digit is equal to or more than the digit precedig it i.e. a ≤ b ≤ c where a, b, c vary from 0 to 9. (Mind that 'a' here can be 0 too)

Now either you can make cases here that all three a, b, c are distinct i.e. in C(10, 3) = 120 ways and can be arranged in one way only.

Or two of a, b, c are same, so two distinct letters can be selected in C(10, 2) = 45 ways and they can be arranged in 2 ways each as which digit is repeating. So it makes 2*45 = 90 more cases.

And lastly all three a, b, c to be idenical i.e. 10 more cases.

So total cases for numbers are = 120 + 90 + 10 = 220. smile

Kamal Lohia    

Re: Permu n Combi - I
by Priyam Garg - Tuesday, 24 May 2016, 06:14 PM
  Hi TG team,
Could you please explain the method to solve q14,16,26,27,and 28.

My answer are
14- couldn't get to it!!
16 - sad
23 - 840,120,24
26 - 2 or 4 do the 3 blocks have to be consecutively placed too?
27 - total confusion
28 - what do we mean by constraint 1&2?

Could someone please confirm these answers.

Thanks
Priyam