In how many ways can the letters of the word PERMUTATIONS be arranged if the

there are always 4 letters between P and S?

# Permutation question

**supriyas**#1

Forget P and S ,lets try to find permuation of other letters remaining ,other left are 10 of which T comes twice.

So 10!/2!.

Now take any permuatation of this say we take

E R M U T A T I O N,

now I have to place P and S such way that there are 4 letters between them.

so I can place P and S as

P ( E R M U) S T A T I O N

or

E P ( R M U T) S A T I ON

similarly i can place P and S between

1 and 6

2 and 7

3 and 8

4 and 9

5 and 10

6 and 11

so 7 places ,also P and S can change its position to so 7* 2 = 14.

so every permutation of without P and S leads to 14 different permutation of there are 4 letters between P and S,

so answer is 14 * 10!/2!

Found out the question has been from ncert 11 maths textbook and asked on gmat forum.

This can also be solved using alternate approach in which you keep P and S fixed.

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

**Kr_Shantanu**#3

There are total 907200 ways.

Total number of letters in PERMUTATIONS is 12 in which T comes 2 times.

Condition given - There should be always 4 letters between P and S.

So P and S are fixed with 4 places between - P_ _ _ _ S. As P and S are fixed, we need to arrange only remaining 10 letters.

So in between P and S possibility is 10P4 which is 10 * 9 * 8 * 7 = 5040. This can also have repeating T, so finally it will be 5040/2 = 2520.

Now P and S along with 4 letters can be considered as 1 so we have remaining 6 places with 6 letters left to arrange with possibility of repetition of T. It will be 6!/2 = 360.

So, conclusive Permutation becomes 2520*360= 907200.

**shreya**#4

The place where you are making mistake is This can also have repeating T, so finally it will be 5040/2 = 2520.You are choosing 4 out of 10,it may happen only 1 T can come,no T can come or 2 T can come,so you cant just divide by 2.

Thanks.