Derangements and the number 1/e

1. Inclusion–exclusion. $D_n=n!-\left|A_1\cup\dots\cup A_n\right|$. For a set $I$ of $k$ indices, the permutations fixing every element of $I$ number $(n-k)!$. There are $\binom nk$ such sets. Therefore
$$D_n=\sum_{k=0}^n(-1)^k\binom nk(n-k)!=\sum_{k=0}^n(-1)^k\frac{n!}{k!}=n!\sum_{k=0}^n\frac{(-1)^k}{k!}.$$

2. Values. $D_4=24-24+12-4+1=9$.

Closest integer. Since $e^{-1}=\sum_{k\ge0}\frac{(-1)^k}{k!}$,
$$\frac{n!}e-D_n=n!\sum_{k>n}\frac{(-1)^k}{k!}.$$
The series is alternating with decreasing terms, so its absolute value is at most the first term: $\left|\frac{n!}e-D_n\right|\le\frac{n!}{(n+1)!}=\frac1{n+1}\le\frac13<\frac12$ for $n\ge2$. So $D_n$ is the integer closest to $\frac{n!}e$.

Consequence. The probability that a random permutation has no fixed point tends to $\frac1e$.

Lesson. Inclusion–exclusion: when the intersections of the bad events all have the same size depending only on the number of events, the sum collapses to a clean formula.