The derangement numbers count derangements

Statement

Let $X$ be a finite set with $|X| = n$. Then the number of derangements of $X$ is the derangement number $D_n$ of the node number of derangements.

Remarks

The proof checks the recursion $D_{n+2} = (n + 1)(D_n + D_{n+1})$, as in the node recurrence for derangements. For $n = 4$ there are $9$ derangements among the $24$ permutations.

Explanations

L1 human

The derangement numbers are defined by a recursion, and this result says that they really count the permutations with no fixed point. For example, $D_3 = 2$, and $\{1, 2, 3\}$ has exactly the two derangements $231$ and $312$.