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permutation and combination problem
by anish verma - Saturday, 20 September 2008, 10:18 AM
  In how many ways can the letters of the word MISSISSIPPI be arranged so that each 'S' is paired with each'I'?sad Help
Re: permutation and combination problem
by sunil garg - Saturday, 20 September 2008, 10:58 AM
 

hi Anish

very good question

is the answer (7!/2!.4!)2^4

regards

sunil garg

Re: permutation and combination problem
by anish verma - Saturday, 20 September 2008, 11:12 AM
  hi Sunil,
iam also searching for solution...can anybody help us
plz....elaborate the concept....thanx.
regards
Anish
Re: permutation and combination problem
by sunil garg - Saturday, 20 September 2008, 12:01 PM
 

CONCEPT

to me .....S & I HAS FOUR PAIRS ( SAY THESE ARE FOUR LETTERS) REST LETTERS ARE M,P,P.....SO WE HAVE NOW 7 LETTERS OUT OF THEM 2 ARE ALIKE AND 4 ARE ALIKE....ALSO IN EACH PAIR S & I CAN BE PUT IN 2 WAYS (SI or IS).....SO THAT FACTOR IS 2^4 ALSO ADDED

WHAT U SAY

Re: permutation and combination problem
by vamsi krishna - Saturday, 20 September 2008, 12:09 PM
 
i feel ans shud be like
 
7! 24  ( 2  +    1    + )
  2!      4!    2!.2!    3!
 
M , P , P , IS, IS , IS , IS
Above seven block can be arranged in 7! WAYS
Divided by 2 ! (coz of two P's)
 
NOW "IS" can be internally arranged as IS or SI  in 24 ways and following 3 cases arise:
1) IS ,IS, IS, IS (1/4!) or SI ,SI, SI, SI  (1/4!)
2) permutations of SI, IS , IS , IS (1/3!) or permutations of IS , SI , SI , SI (1/3!)
3) permutations of SI, SI , IS , IS (1/(2!.2!))
 
Hence
7! 24  ( 2  +    1    + )
  2!      4!    2!.2!    3!
 
i would be grateful if Someone could Comment on this approach
 
VaMsI
Re: permutation and combination problem
by anish verma - Saturday, 20 September 2008, 12:50 PM
  I think ur approach is correct . Because four SI looks similar. But actually it is only one case There are other cases as well which u have considered. But I think TG guru should clarify this question and give us a firm answer . Thank u.

TG help...
Regards
Anish
Re: permutation and combination problem
by sunil garg - Saturday, 20 September 2008, 02:13 PM
 

HI VAMSI

UR Approach same on line ....in fact better.....

but if u r taking those cases different then i dont think we need ta take factor of 2^4.......

WHAT u say little bit more to think ...i need my ciggy to think....

regards

sunil

Re: permutation and combination problem
by vamsi krishna - Saturday, 20 September 2008, 06:56 PM
  I 2 feel that itz not necessary , but am workin on tat...

VaMsI
Re: permutation and combination problem
by aarti kaushik - Sunday, 21 September 2008, 01:02 PM
 

Plz solve my problem.

Q.1 The number of permutations of the letters a,b,c,d,e,f,g  such tht neither the pattern 'beg' nor 'acd' occurs is:

a)4806 b)420 c)2408 d)140

Q.2 The number of cicles that drawn out of 10 points of which 7 are collinear is ?????????????

 

 

 

Re: permutation and combination problem
by bimal mohan - Sunday, 21 September 2008, 08:09 PM
 

Dear  all,

        MISSISSIPPI   :   S ---4 ,  I ----  4, M---1, P----2

as  S   and  I  will  stay  in   pair   so  the  combination  will  be  either  '--SI--- ' or '----IS----'

Let   '--SI--- '  be  X   and  '----IS----'  be  Y    then  in  diff  words  these   may   come   as :

   X ---4/3/2/1/0

  Y----0/1/2/3/4  ( correspondingly)

  M----1

  P --- 2

    so,  reqd   value  =  [ 7!/4!*2! +7!/3!*2! +7!/2!*2!*2! +7!/3!*2! +7!/4!*2!]

                            =  7!/2![2/4!+2/3!+1/2!*2!]

pls  suggest   if  i'm  wrong.

Bimal

Re: permutation and combination problem
by Floydian _iva - Sunday, 21 September 2008, 09:36 PM
  1. The answer is 7! - 3! -2[ 2(4!-1) + 4! +2(4!-2)] = 4806 (b)
Approach: total number of words: 7!
                Number with both the patters: 3!
                Number with only one pattern acd: 2(4!-1) + 4! +2(4!-2)
                Number with only one pattern beg: 2(4!-1) + 4! +2(4!-2)
               
 Regards
 Avi              
        
Re: permutation and combination problem
by Floydian _iva - Sunday, 21 September 2008, 10:26 PM
  2. 3C3 + 7C1*3C2 + 7C2*3C1 = 43
    I am not very sure. Please confirm.
    Regards
    Avi
   
Re: permutation and combination problem
by sunil garg - Sunday, 21 September 2008, 10:30 PM
 

yarro

pehle MISSISSIPI to solve karo

 

Re: permutation and combination problem
by Floydian _iva - Sunday, 21 September 2008, 11:10 PM
  Sunil
Don't smoke. It's like you are paying money to kill yourself, that too in instalment.
Well, the answer will be: 7!/2!( 2/4! + 1/2!*2! + 2/3!).
Vamsi's approach is perfect expect the factor 2^4 won't be there. It's like double counting, we have already considered those cases.
Regards
Avi.

Re: permutation and combination problem
by sunil garg - Sunday, 21 September 2008, 11:51 PM
 

thanx buddy

for ur sincere advice............ya that answer seems logical

gr8 keep it up....same type qn appeared today in SIMCAT10

Re: permutation and combination problem
by vamsi krishna - Monday, 22 September 2008, 10:11 AM
 

Hey avi 'n' Bimal

Yep ther shudnt be tat 2^ 4 thing...

Thanks

Re: permutation and combination problem
by vamsi krishna - Monday, 22 September 2008, 11:01 AM
 
1)
7! -(5!+ 5! - 3!) = 4806
ans = total permutations - ('acd' cases +'beg' cases -'acd' & 'beg' cases)
TOtal permutations = 7!
acd pattern  = 5!
beg pattern = 5!
both acd and beg pattern 3 !
 
isnt this much simpler sir!
 
2)
I feel we cant arrive at any particular number( may be 1 or more of the 3 non-collinear points is/are lying on one or more circles formed by any other three points, we never know)
 
Assuming the case where no four points are on same circle we can arrive at Max. Number of circles as  C(10,3) - C(7,3)
 
Commentz solicited
 
VaMsI
Re: permutation and combination problem
by Floydian _iva - Monday, 22 September 2008, 12:21 PM
  Vamsi,
That was an example how a simple thing can be done in a complicated way. Anyway...thank you very much.
For the second one, I am not very sure about the answer. I could not even understand your explanation.
Regards
Avi
Re: permutation and combination problem
by vamsi krishna - Monday, 22 September 2008, 01:56 PM
  Hey avi
Consider this!
How many circles can be drawn thru 4 non-collinear , coplanar points?

4c3 = 4??

huh..may be
Depends on da location of points in plane
max->4 circles , min->1 circle..

hope u got my intention!

VaMsI
Re: permutation and combination problem
by aarti kaushik - Monday, 22 September 2008, 07:55 PM
 

Vamsi,

The options for 2nd question is

a)130 b)85 c)45 d)72

Yes, 85 is correct answer.But i m nt getting ur approach.

Try to explain it  more????????????

 

 

Re: permutation and combination problem
by vamsi krishna - Monday, 22 September 2008, 08:24 PM
 
Yep,aarti

2)
my point is that 85 is max.numbers of circles possible, and there may be cases where number of circles is less than that...

consider simple case below..where arrangement of noncollinear points also decides the number of circles possible..

Re: permutation and combination problem
by Ravi Shankar - Tuesday, 23 September 2008, 03:53 PM
  Are yaar MISSISSIPPI ka to answer batao pehle ...I mean Final Answer
Re: permutation and combination problem
by Yogesh Kumar - Thursday, 14 January 2016, 02:51 AM
  while considering acd, can it be possible to put any alphabet in between them? e.g a"b"cd. b in in between acd..