Recurrence for derangements

Builds on Derangement Number of derangements

Statement

Let $n$ be a natural number, and let $D_m$ be the number of derangements of an $m$ -element set. Then $D_{n+2} = (n + 1)\,(D_{n+1} + D_n).$

Remarks

Together with $D_0 = 1$ and $D_1 = 0$ , the recurrence gives all values: $1, 0, 1, 2, 9, 44, 265, \dots$ A proof looks at a derangement $\sigma$ of $\{1, \dots, n + 2\}$ and the value $i = \sigma(n+2)$ . There are $n + 1$ choices of $i$ . The case $\sigma(i) = n + 2$ gives $D_n$ derangements, and the other case gives $D_{n+1}$ . Another recurrence is $D_{n+1} = (n + 1) D_n + (-1)^{n+1}$ .

Reading path

To understand this page from the ground up, read these 4 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

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Number of derangements definition
Permutation definition
Derangement definition
Recurrence for derangements statement

Tag	`0QRRYXG7V7`
URI	`m2://0QRRYXG7V7`
Kind	statement, theorem
Writing	W0 stub
MSC	05A05
Status	This copy of the site has no build data. Run `m2 status 0QRRYXG7V7`.
Uses	Derangement, Number of derangements

Recurrence for derangements

Statement

Remarks

Reading path

Further reading

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