In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position. In other words, a derangement is a permutation that has no fixed points.

The number of derangements of a set of size n is known as the subfactorial of n or the n-th derangement number or n-th de Montmort number (after Pierre Remond de Montmort). Notations for subfactorials in common use include !n, D_n, d_n, or n¡.^[1][2]

For n > 0, the subfactorial !n equals the nearest integer to n!/e, where n! denotes the factorial of n and e is Euler's number.^[3]

The problem of counting derangements was first considered by Pierre Raymond de Montmort in his Essay d'analyse sur les jeux de hazard.^[4] in 1708; he solved it in 1713, as did Nicholas Bernoulli at about the same time.

Example

Suppose that a professor gave a test to 4 students – A, B, C, and D – and wants to let them grade each other's tests. Of course, no student should grade their own test. How many ways could the professor hand the tests back to the students for grading, such that no student received their own test back? Out of 24 possible permutations (4!) for handing back the tests,

ABCD,	ABDC,	ACBD,	ACDB,	ADBC,	ADCB,
BACD,	BADC,	BCAD,	BCDA,	BDAC,	BDCA,
CABD,	CADB,	CBAD,	CBDA,	CDAB,	CDBA,
DABC,	DACB,	DBAC,	DBCA,	DCAB,	DCBA.

there are only 9 derangements (shown in blue italics above). In every other permutation of this 4-member set, at least one student gets their own test back (shown in bold red).

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.

Counting derangements

Counting derangements of a set amounts to the hat-check problem, in which one considers the number of ways in which n hats (call them h₁ through h_n) can be returned to n people (P₁ through P_n) such that no hat makes it back to its owner.^[5]

Each person may receive any of the n − 1 hats that is not their own. Call the hat which the person P₁ receives h_i and consider h_i's owner: P_i receives either P₁'s hat, h₁, or some other. Accordingly, the problem splits into two possible cases:

1. P_i receives a hat other than h₁. This case is equivalent to solving the problem with n − 1 people and n − 1 hats because for each of the n − 1 people besides P₁ there is exactly one hat from among the remaining n − 1 hats that they may not receive (for any P_j besides P_i, the unreceivable hat is h_j, while for P_i it is h₁). Another way to see this is to rename h₁ to h_i, where the derangement is more explicit: for any j from 2 to n, P_j cannot receive h_j.
2. P_i receives h₁. In this case the problem reduces to n − 2 people and n − 2 hats, because P₁ received h_i's hat and P_i received h₁'s hat, effectively putting both out of further consideration.

For each of the n − 1 hats that P₁ may receive, the number of ways that P₂, ..., P_n may all receive hats is the sum of the counts for the two cases.

This gives us the solution to the hat-check problem: stated algebraically, the number !n of derangements of an n-element set is

${\displaystyle !n=(n-1)\cdot ({!(n-1)}+{!(n-2)})}$ for ${\displaystyle n\geq 2}$,

where ${\displaystyle !0=1}$ and ${\displaystyle !1=0}$.^[6]

The number of derangements of small lengths is given in the table below.

There are various other expressions for !n, equivalent to the formula given above. These include

${\displaystyle !n=n!\sum _{i=0}^{n}{\frac {(-1)^{i)){i!))}$ for ${\displaystyle n\geq 0}$

and

${\displaystyle !n=\left[{\frac {n!}{e))\right]=\left\lfloor {\frac {n!}{e))+{\frac {1}{2))\right\rfloor }$ for ${\displaystyle n\geq 1,}$

where ${\displaystyle \left[x\right]}$ is the nearest integer function and ${\displaystyle \left\lfloor x\right\rfloor }$ is the floor function.^[3][6]

Other related formulas include^[3][7]

$${\displaystyle !n=\left\lfloor {\frac {n!+1}{e))\right\rfloor ,\quad \ n\geq 1,}$$

$${\displaystyle !n=\left\lfloor \left(e+e^{-1}\right)n!\right\rfloor -\lfloor en!\rfloor ,\quad n\geq 2,}$$

$${\displaystyle !n=n!-\sum _{i=1}^{n}{n \choose i}\cdot {!(n-i)},\quad \ n\geq 1.}$$

The following recurrence also holds:^[6]

$${\displaystyle !n={\begin{cases}1&{\text{if ))n=0,\\n\cdot \left(!(n-1)\right)+(-1)^{n}&{\text{if ))n>0.\end{cases))}$$

Derivation by inclusion–exclusion principle

One may derive a non-recursive formula for the number of derangements of an n-set, as well. For ${\displaystyle 1\leq k\leq n}$ we define ${\displaystyle S_{k))$ to be the set of permutations of n objects that fix the ${\displaystyle k}$-th object. Any intersection of a collection of i of these sets fixes a particular set of i objects and therefore contains ${\displaystyle (n-i)!}$ permutations. There are ${\textstyle {n \choose i))$ such collections, so the inclusion–exclusion principle yields

$${\displaystyle {\begin{aligned}|S_{1}\cup \dotsm \cup S_{n}|&=\sum _{i}\left|S_{i}\right|-\sum _{i<j}\left|S_{i}\cap S_{j}\right|+\sum _{i<j<k}\left|S_{i}\cap S_{j}\cap S_{k}\right|+\cdots +(-1)^{n+1}\left|S_{1}\cap \dotsm \cap S_{n}\right|\\&={n \choose 1}(n-1)!-{n \choose 2}(n-2)!+{n \choose 3}(n-3)!-\cdots +(-1)^{n+1}{n \choose n}0!\\&=\sum _{i=1}^{n}(-1)^{i+1}{n \choose i}(n-i)!\\&=n!\ \sum _{i=1}^{n}{(-1)^{i+1} \over i!},\end{aligned))}$$

$${\displaystyle !n=n!-\left|S_{1}\cup \dotsm \cup S_{n}\right|=n!\sum _{i=0}^{n}{\frac {(-1)^{i)){i!)).}$$

On the other hand, ${\displaystyle n!=\sum _{i=0}^{n}{\binom {n}{i))!i}$ since we can choose n - i elements to be in their own place and derange the other i elements in just !i ways, by definition.^[8]

Growth of number of derangements as n approaches ∞

From

$${\displaystyle !n=n!\sum _{i=0}^{n}{\frac {(-1)^{i)){i!))}$$

$${\displaystyle e^{x}=\sum _{i=0}^{\infty }{x^{i} \over i!))$$

${\displaystyle \textstyle x=-1}$

$${\displaystyle \lim _{n\to \infty }{!n \over n!}=\lim _{n\to \infty }\sum _{i=0}^{n}{\frac {(-1)^{i)){i!))=e^{-1}\approx 0.367879\ldots .}$$

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

Asymptotic expansion in terms of Bell numbers

An asymptotic expansion for the number of derangements in terms of Bell numbers is as follows:

$${\displaystyle !n={\frac {n!}{e))+\sum _{k=1}^{m}\left(-1\right)^{n+k-1}{\frac {B_{k)){n^{k))}+O\left({\frac {1}{n^{m+1))}\right),}$$

${\displaystyle m}$

${\displaystyle B_{k))$

${\displaystyle k}$

${\displaystyle B_{m+1))$

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 opposite-sex couples are seated man-woman-man-woman-... around a table, how many ways can they be seated so that nobody is seated next to his or her partner?

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)).

Another generalization is the following problem:

How many anagrams with no fixed letters of a given word are there?

For instance, for a word made of only two different letters, say n letters A and m letters B, the answer is, of course, 1 or 0 according to whether n = m or not, for the only way to form an anagram without fixed letters is to exchange all the A with B, which is possible if and only if n = m. In the general case, for a word with n₁ letters X₁, n₂ letters X₂, ..., n_r letters X_r, it turns out (after a proper use of the inclusion-exclusion formula) that the answer has the form

$${\displaystyle \int _{0}^{\infty }P_{n_{1))(x)P_{n_{2))(x)\cdots P_{n_{r))(x)e^{-x}\,dx,}$$

In particular, for the classical derangements, one has that

$${\displaystyle !n={\frac {\Gamma (n+1,-1)}{e))=\int _{0}^{\infty }(x-1)^{n}e^{-x}dx}$$

${\displaystyle \Gamma (s,x)}$

Computational complexity

It is NP-complete to determine whether a given permutation group (described by a given set of permutations that generate it) contains any derangements.^[11]