Closed Form for Number of Derangements on Finite Set/Proof 1

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 permutations, $W$, with at least one element fixed, $m$, is:
 * $\ds W = \size {\bigcup_{m \mathop = 1}^n A_m}$

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 $\dbinom n p$ terms in each summation.

So:

$\size {A_1 \cap \cdots \cap A_p}$ is the number of permutations fixing $p$ elements in position.

This is equal to the number of permutations which rearrange the remaining $n - p$ elements, which is $\paren {n - p}!$.

Thus we finally get:


 * $W = \dbinom n 1 \paren {n - 1}! - \dbinom n 2 \paren {n - 2}! + \dbinom n 3 \paren {n - 3}! - \cdots + \paren {-1}^{p - 1} \dbinom n p \paren {n - p}! \cdots$

That is:


 * $\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \binom n p \paren {n - p}!$

Noting that $\dbinom n p = \dfrac {n!} {p! \paren {n - p}!}$, this reduces to:


 * $\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \dfrac {n!} {p!}$