Closed Form for Number of Derangements on Finite Set

Theorem
The number of derangements on a finite set $S$ where $\left|{S}\right| = n$ is:


 * $\displaystyle D_n = n! \left({1 - \frac 1 {1!} + \frac 1 {2!} - \frac 1 {3!} + \cdots + \left({-1}\right)^n \frac 1 {n!} }\right)$

Proof
Let $s_i$ be the $i$th element of set $S$.

Begin by defining set $A_m$, which is all of the permutations of $S$ which fixes $S_m$.

Then the number of orders, $W$, with at least one element fixed, m, is:
 * $\displaystyle W = \left|{\bigcup_{m=1}^n A_m}\right|$

Applying the Inclusion-Exclusion Principle:

Each value $A_{m_1} \cap \cdots \cap A_{m_p}$ represents the set of permutations which fix $p$ values $m_1, \ldots, m_p$.

Note that the number of permutations which fix $p$ values only depends on $p$, not on the particular values of $m$.

Thus from Cardinality of Set of Subsets there are $\displaystyle \binom n p$ terms in each summation.

So:

$\left|{A_1 \cap \cdots \cap A_p} \right|$ is the number of permutations fixing $p$ elements in the correct position, which is equal to the number of permuting the remaining $n - p$ elements, or $\left({n - p}\right)!$.

Thus we finally get:


 * $\displaystyle W = \binom n 1 (n-1)! - \binom n 2 (n-2)! + \binom n 3 (n-3)! - \cdots + \left({-1}\right)^{p-1} \binom n p (n-p)! \cdots $

That is:


 * $\displaystyle W = \sum_{p=1}^n (-1)^{p-1} \binom n p (n-p)!$

Noting that $\displaystyle \binom n p = \frac{n!}{p!(n-p)!}$, this reduces to:
 * $\displaystyle W = \sum_{p=1}^n (-1)^{p-1} \frac{n!}{p!}$