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