How many arrangements can we make from the letters from the word PERMUTATIONS such that there are always 4 letters in between P and S?

I think it is ((superscript)10P(subscript)4)/2! * 6! because since P and S are utilised there are only 10 letters left and , since only 4 letters should be there in between P and S, there are 4 slots, so, the number of possible arrangements between P and S is 10*9*8*7 but, T repeats twice...so it hsuld be 10P4/2! and, this "object" between P and S can be arranged in 6 ways within the huge word...

But, according to the book the answer isn't matching...it starts with 2.......000, sorry, don't have the answer list, lost it

, but still. please help..