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I figured out the solution however it is not a simple permutation question. This includes both permutation as well as combination.

There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice.

Now first we need to see how many ways we can make word with 4 letter between P and S. Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210

Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S. So total way = 210*2 = 420

The selected 4 letters can be rotated between P and S in = 4! ways

So total ways = 420 * 4!

Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter. Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways.

So total number of ways = 7! * 420 * 4!

Now since letter T was repeated twice, we should divide the above result by 2!.

So Total number of ways = 7! * 420 * 4! / 2! = 25401600

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

. . . . . _ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

. . . . . _ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

Thanks

Ravi

Thanks! man. kudos
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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10 Nov 2013, 03:30

Hello from the GMAT Club BumpBot!

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Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

Back to the original questions:

In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?

There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice.

1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210; 2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2; 3. Permutation of the 4 letters between P and S = 4! =24; 4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040; 5. We should divide multiplication of the above 4 numbers by 2! as there is repeated T.

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