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Section 3.3 Application: Derangements

Video: Derangements

Recall that the permutations of a set S are the bijective functions from S to itself. We have a special name for those permutations that leave no element fixed:

Definition 3.3.1. Derangement.

A derangement is a permutation on {1,2,,n} such that no element is mapped to itself.

The permutation on \(\{1,2,3\}\) that takes \(1\) to \(3\text{,}\) \(2\) to \(1\text{,}\) and \(3\) to \(2\) is a derangement; we can also denote it as the string 312.

There are only two derangements on \(\{1,2,3\}\text{:}\) 231 and 312.

The permutation 3241 on \(\{1,2,3,4\}\) is not a derangement since 2 is sent to itself. We call 2 a fixed point of the permutation.

(Try listing all derangements of \(\{1,2,3,4\}\text{.}\))

Exploration 3.3.1. Deriving Dn.

Count the number of derangements Dn of the set {1,2,,n} using Theorem 3.2.6, by following these steps:

(a)

Define Ai to be the set of permutations of {1,2,,n} for which i is a fixed point. (There are no restrictions on the other elements; there may be other fixed points.)

Then, express Dn as the cardinality of a set involving the Ai's.

(b)

If S{1,2,,n} with |S|=k, find a formula for |iSAi|.

(c)

Combine with Theorem 3.2.6 to derive a formula for Dn, then fill in the statement of the result below.

Remark 3.3.4.

For problems where it is relevant, you may leave your answers in terms of Dn. (You don't have to compute exact values unless asked.)

Evaluate \(D_4\) and verify that it is the correct number by listing all derangements of \(\{1,2,3,4\}\text{.}\)

Using a computer, evaluate the ratio

\begin{equation*} \frac{D_n}{n!} \end{equation*}

of derangements to all permutations of \(\{1,2,\ldots,n\}\) for \(n\) increasingly large.

Verify that the ratios approach the value of \(\frac{1}{e}\text{.}\)