# Closed Form for Number of Derangements on Finite Set

## Theorem

The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:

$D_n = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }$

## 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:

$\displaystyle W = \size {\bigcup_{m \mathop = 1}^n A_m}$

Applying the Inclusion-Exclusion Principle:

 $\displaystyle W$ $=$ $\displaystyle \sum_{m_1 \mathop = 1}^n \size {A_{m_1} }$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{m_1, m_2 : 1 \mathop \le m_1 \mathop < m_2 \mathop \le n} \size {A_{m_1} \cap A_{m_2} }$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{m_1, m_2, m_3 : 1 \mathop \le m_1 \mathop < m_2 \mathop < m_3 \mathop \le n} \size {A_{m_1} \cap A_{m_2} \cap A_{m_3} }$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \cdots$

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:

 $\displaystyle W$ $=$ $\displaystyle \binom n 1 \size {A_1}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \binom n 2 \size {A_1 \cap A_2}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \binom n 3 \size {A_1 \cap A_2 \cap A_3}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \cdots$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{p - 1} \binom n p \size {A_1 \cap \cdots \cap A_p}$ $\displaystyle$  $\displaystyle \cdots$

$\size {A_1 \cap \cdots \cap A_p}$ 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:

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

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

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

$\blacksquare$

## Historical Note

This result was first solved by Nicolaus I Bernoulli.

Leonhard Paul Euler later solved it independently.