You have $n-1$ options to choose the value of $f(1)$, it can't be $1$. Then you have $n-2$ options to choose $f^2(1)$, because it can't be $1$ and can't be $f(1)$. Then $n-3$ options to choose the value of $f^3(1)$. Continue this way and you will get the number of $n$-cycles is $(n-1)!$.
No, your answer is not correct.
The goal is to count the number of permutations $p\in S_{2n}$ such that in the disjoint cycle representation of $p$, the maximum cycle length is $n$.
You need to worry about the case where $p$ is a product of two disjoint $n$-cycles.
Thus, consider two cases . . .
Case $(1)$:$\;$The disjoint cycle representation of $p$ has only one cycle of length $n$.
For case $(1)$, the count is
$$\binom{2n}{n}{\,\cdot\,}(n-1)!{\,\cdot\,}\bigl(n!-(n-1)!\bigr)$$
Explanation:
- The factor ${\large{\binom{2n}{n}}}$ counts the $n$-element subsets of $\{1,...,2n\}$ used to form the $n$-cycle.$\\[4pt]$
- The factor $(n-1)!$ counts the cyclic orderings of the $n$ elements used for the $n$-cycle.$\\[4pt]$
- The factor $n!-(n-1)!$ counts the $n!$ permutations of the remaining $n$ elements, but excludes the $(n-1)!$ permutations that would form an $n$-cycle.
Case $(2)$:$\;p$ is a product of two disjoint $n$-cycles.
For case $(2)$, the count is
$$\binom{2n}{n}{\,\cdot\,}\bigl((n-1)!\bigr)^2{\,\cdot\,}\bigl({\small{\frac{1}{2}}}\bigr)$$
Explanation:
- The factor ${\large{\binom{2n}{n}}}$ counts the $n$-element subsets of $\{1,...,2n\}$ used to form the first $n$-cycle.$\\[4pt]$
- The factor $\bigl((n-1)!\bigr)^2$ counts the cyclic orderings of the elements for the two $n$-cycles.$\\[4pt]$
- The factor ${\large{\frac{1}{2}}}$ corrects for the double-count, since we can freely switch the order of the two $n$-cycles, without affecting the result.
Summing the counts for the two cases, and then simplifying, we get a total count of
$$
(2n-1)!{\;\cdot}\left(\frac{2n-1}{n}\right)
$$
Best Answer
For the cycle of length $4$, you have $9$ choices for the first element, $8$ for the second, $7$ for the third, and $6$ for the fourth. But you have to divide by $4$ because for example $(1234)=(2341)$.
For the cycle of length $3$ you then have $5$ choices for the first element, $4$ for the second, and $3$ for the third; divide by $3$ because there are three ways to represent the same cycle.
Once you have chosen the cycles of length $4$ and $3$, the cycle of length $2$ is uniquely determined. (There are two possibilities for the first element, but the cycle is the same no matter which of the remaining two elements is written first.)