Derangement

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In combinatorial mathematics, a derangement is a permutation φ of a set S (i.e., a bijection from S into itself) with no fixed points, i.e., for all x in S, φ(x) ≠ x. A frequent problem is to count the number of derangements as a function of n = |S|, often with additional constraints.

Here is a concrete example: an instructor returns graded exams to students named "A", "B", "C", and "D", but neglects to assure that each student gets his or her own test. In how many ways can this be done without any student getting the right paper back? There are 9:

BADC, BCDA, BDAC,

CADB, CDAB, CDBA,

DABC, DCAB, DCBA

In all of the other 15 permutations of this 4-member set, at least one student gets the right paper.

Another version of the problem arises when we ask for the number of ways n letters, each addressed to a different person, can be placed in n pre-addressed envelopes so that no letter appears in the correctly addressed envelope.

One approach to counting the derangements of n elements is to use induction. First, note that if φ_n is any derangement of the natural numbers , then for some k in , φ_n(n) = k. Then if we let (k, n) be the permutation of which swaps k and n, and we let φ_n − 1 be the composition ((k, n) o φ_n); then φ_n−1(n) = n, and either:

φ_n − 1(k) ≠ k, so φ_n − 1 is a derangement of , or
φ_n−1(k) = k, and for all x ≠ k, φ_n−1(x) ≠ x.

In the latter case, φ_n − 1 is then a derangement of the set excluding k; i.e., the composition φ_n−2 = ((k,n − 1) o φ_n − 1 o (k,n−1)) is a derangement of .

As examples of these two cases, consider the following two derangements of 6 elements as we perform the above described swaps:

514623 → (51432)6; and

315624 → (31542)6 → (3142)56

The above described correspondences are 1-to-1. The converse is also true; there are exactly (n − 1) ways of converting any derangement of n − 1 elements into a derangement of n elements, and (n − 1) ways of converting any derangement of n − 2 elements into a derangement of n elements. For example, if n = 6 and k = 4, we can perform the following conversions of derangements of length 5 and 4, respectively

51432 → 514326 → 514623; and

3142 → 31542 → 315426 → 315624

Thus, if we write d_n as the number of derangements of n letters, and we define d₀ = 1, d₁ = 0; then d_n satisfies the recurrence:

[d_n = (n - 1) (d_ + d_)\,]

Limit as n approaches ∞

Using this recurrence, it can be shown that, in the limit,

[\lim_ = \approx 0.3679\dots.]

This is the limit of the probability p_n = d_n/n! that a randomly selected permutation is a derangement. The probability approaches this limit quite quickly.

Perhaps a more well-known method of counting derangements uses the inclusion-exclusion principle.

Generalizations

The problème des rencontres asks how many permutations of a size-n set have exactly k fixed points.

Derangements are an example of the wider field of constrained permutations. For example, the ménage problem asks if n married couples are seated boy-girl-boy-girl-... around a circular table, how many ways can they be seated so that no man is seated next to his wife?

More formally, given sets A and S, and some sets U and V of surjections A → S, we often wish to know the number of pairs of functions (f,g) such that f is in U and g is in V, and for all a in A, f(a) ≠ g(a); in other words, where for each f and g, there exists a derangement φ of S such that f(a) = φ(g(a)).

External links

Sloan's Integer Sequence [A000166]
[Non-sexist solution of the ménage problem], Kenneth P. Bogart, Peter G. Doyle

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