Limiting chance a random permutation fixes five points

Let $F_n$ denote the number of fixed points of a permutation of $\{1,\dots,n\}$ chosen uniformly at random. The limit of $\mathbb{P}(F_n=5)$ as $n\to\infty$ equals $c/e$. Find $c$.

$c=\tfrac1{120}$. The number of fixed points of a uniform random permutation converges in distribution to $\text{Poisson}(1)$. Inclusion–exclusion gives this exactly: $\mathbb{P}(F_n=k)=\tfrac1{k!}\sum_{j=0}^{n-k}\tfrac{(-1)^j}{j!}\to\tfrac{e^{-1}}{k!}$, so the limit is $\tfrac{1}{120e}$.