In how many ways can the word "permutations" be arranged if there are 4 letters between p and s?

Question

In how many ways can the word "permutations" be arranged if there are 4 letters between p and s?

I simply can't get any hold on this problem

Answer 1

Let p,s be at $1,6$ positions respectively.Now total ways of arranging remaining letters is $\frac{10!}{2!} $. Now $p,s$ can be at $(2,7), (3,8), (4,9), (5,10), (6,11), (7,12) $. Also p and s can be arranged in $2! $ ways hence total ways are $2!.7.\frac {10!}{2!}=7.10! $