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

This is a 12 letter word. Let each position of the word be numbered. So there are 12 positions to fill in. _ _ _ _ _ _ _ _ _ _ _ _ if P is in position 1 then 'S' will be in number 6, position 2,3,4,5 will be occupied by some other four letters. Let us call this pair (1,6) Similarly for position of 'P' at 2 ,3 4, 5 ,6 and 7. This gives 7 cases. If we reverse order between p and s we get another 7 cases Total 14 cases. Leaving out p and s there are 10 letters with t coming twice. These can take any of those 10 places left out by gaps in 10!/2! ways. So total 14* 10!/2! = 7 * 10*9*8*7*6*5*4*3*2 = 25401600 read less

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 read less