Finding all permutations which satisfy given condition

Question

In a symmetric group $S_n$ find number of permutations $P$ such that in the disjoint cycle decomposition of $P$ , length of cycle containing $1$ is $k$ . Here's my attempt at this . I found number of cycles containing $1$ and of length $k$ as $\frac{(n-1)!}{(n-k)!}$. Rest of them can be permutated in $(n-k)!$ ways. So number of such permutations $P$ comes out to be $(n-1)!$. But I doubt my answer because it is independent of $k$. Please help me if I proceeded right or wrong!

Answer 1

The solution is correct, the number of permutations with a fixed length of the cycle containing 1 is $(n-1)!$. The length may be any number between $1$ and $n$, i.e. $n$ choices, so the total number of permutations is $n\times (n-1)!=n!$, as expected.

Because this $(n-1)!$ is independent of $k$, we may say that all the lengths of the cycle including $1$ are "equally likely" among the permutations.