Formula for the number of derangements

Builds on Factorial Derangement Number of derangements

Statement

Let $n$ be a natural number, and let $D_n$ be the number of derangements of an $n$ -element set. Then $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} = \sum_{k=0}^{n} (-1)^k \frac{n!}{k!}.$

Remarks

For $n = 4$ : $24 - 24 + 12 - 4 + 1 = 9$ . The proof applies the inclusion–exclusion principle to the sets $A_i$ of permutations that fix $i$ . The sum is a partial sum of the series of $e^{-1}$ . Thus $D_n / n! \to 1/e$ , and for $n \ge 1$ , $D_n$ is the integer nearest to $n!/e$ .

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Factorial definition
Number of derangements definition
Permutation definition
Derangement definition
Formula for the number of derangements statement

Tag	`03RSJXF728`
URI	`m2://03RSJXF728`
Kind	statement, theorem
Writing	W0 stub
MSC	05A05
Status	This copy of the site has no build data. Run `m2 status 03RSJXF728`.
Uses	Factorial, Derangement, Number of derangements

Formula for the number of derangements

Statement

Remarks

Reading path

Further reading

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