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
$$
(2n-1)!{\;\cdot}\left(\frac{2n-1}{n}\right)
$$
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.)
Best Answer
There are $\dbinom{20}{11}$ ways to choose a set of $11$ elements out of $20$ that will form the longest cycle. Once this set is chosen, there are $(11-1)!=10!$ different ways to arrange these $11$ elements in a cycle. Finally, there are $(20-11)!=9!$ ways to permute the remaining $9$ elements.
Therefore, there are $\dbinom{20}{11}\cdot10!\cdot9!$ ways.