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In combinatorial mathematics, a derangement is a permutation that leaves no element unpermuted. That is, it is a bijection φ from a set S into itself with no fixed points: for all x in S, φ(x) ≠ x. A frequent problem is to count the number of derangements as a function of the number of elements of the set, often with additional constraints; these numbers are called subfactorials and are a special case of the rencontres numbers. The problem of counting derangements was first considered by Pierre Raymond de Montmort in 1708; he solved it in 1713, as did Nicholas Bernoulli at about the same time.

Contents 1 Example 2 Counting derangements 3 Limit as n approaches ∞ 4 Generalizations 5 References 6 External links

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 { 1, ..., n }, then for some k in { 1, ..., n − 1 }, φ_n(n) = k. Then if we let (k, n) be the permutation of { 1, ..., n } 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 { 1, ..., n − 1 }, 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 { 1, ..., n − 1 } excluding k; i.e., the composition φ_n−2 = ((k,n − 1) o φ_n − 1 o (k,n−1)) is a derangement of { 1, ..., n − 2 }.

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:

and also

Notice that this same recurrence formula also works for factorials with different starting values. That is 0! = 1, 1! = 1 and

which is helpful in proving the relationship with e below.

Starting with n = 0, the numbers of derangements, d_n, are:

1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961, 14684570, 176214841, 2290792932, ... (sequence A000166 in OEIS).

These numbers are also called subfactorial or rencontres numbers.

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

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.

More information about this calculation and the above limit may be found on the page on the statistics of random permutations.

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)).

• de Montmort, P. R. (1708). Essay d'analyse sur les jeux de hazard. Paris: Jacque Quillau. Seconde Edition, Revue & augmentée de plusieurs Lettres. Paris: Jacque Quillau. 1713.

• Baez, John (2003). Let's get deranged!.

• Bogart, Kenneth P. and Doyle, Peter G. (1985). Non-sexist solution of the ménage problem.

• Dickau, Robert M.. Derangement diagrams. Figures Using Mathematica.

• Weisstein, Eric W. Derangement. MathWorld–A Wolfram Web Resource.