# Derangement Number of possible permutations and derangements of n elements. n! (n factorial) is the number of n-permutations; !n (n subfactorial) is the number of derangements — n-permutations where all of the n elements change their initial places.

In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that 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. Notations for subfactorials in common use include !n, Dn, dn, or n¡.

One can show that !n equals the nearest integer to n!/e, where n! denotes the factorial of n and e is Euler's number.

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.

## 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 his or her own test. How many ways could the professor hand the tests back to the students for grading, such that no student received his or her own test back? Out of 24 possible permutations (4!) for handing back the tests:

there are only 9 derangements (shown in blue italics above). In every other permutation of this 4-member set, at least one student gets his or her 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

Suppose that there are n people who are numbered 1, 2, ..., n. Let there be n hats also numbered 1, 2, ..., n. We have to find the number of ways in which no one gets the hat having the same number as their number. Let us assume that the first person takes hat i. There are n − 1 ways for the first person to make such a choice. There are now two possibilities, depending on whether or not person i takes hat 1 in return:

1. Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons and n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).
2. Person i takes the hat 1. Now the problem reduces to n − 2 persons and n − 2 hats.

From this, the following relation is derived:

$!n=(n-1)(!(n-1)+!(n-2)).\,$ where !n, known as the subfactorial, represents the number of derangements, with the starting values !0 = 1 and !1 = 0.

Also, the following equalities are known:

$!n=n!\sum _{i=0}^{n}{\frac {(-1)^{i}}{i!}},$ $!n=\left[{\frac {n!}{e}}\right]=\left\lfloor {\frac {n!}{e}}+{\frac {1}{2}}\right\rfloor ,\quad n\geq 1$ where $\left[x\right]$ is the nearest integer function and $\left\lfloor x\right\rfloor$ is the floor function.

$!n=\left\lfloor (e+e^{-1})n!\right\rfloor -\lfloor en!\rfloor ,\quad n\geq 2,$ $!n=n!-\sum _{i=1}^{n}{n \choose i}\cdot !(n-i),$ The following recurrence equality also holds:

$!n=n\left(!(n-1)\right)+(-1)^{n}$ Starting with n = 0, the numbers of derangements of n are:

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

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

{\begin{aligned}|S_{1}\cup \cdots \cup S_{n}|&={n \choose 1}(n-1)!-{n \choose 2}(n-2)!+{n \choose 3}(n-3)!-\cdots \\&=\sum _{i=1}^{n}(-1)^{i+1}{n \choose i}(n-i)!\\&=n!\ \sum _{i=1}^{n}{(-1)^{i+1} \over i!}\end{aligned}} and since a derangement is a permutation that leaves none of the n objects fixed, we get

$!n=n!-|S_{1}\cup \cdots \cup S_{n}|=n!\sum _{i=0}^{n}{\frac {(-1)^{i}}{i!}}.$ ## Limit of ratio of derangement to permutation as n approaches ∞

From

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

$e^{x}=\sum _{i=0}^{\infty }{x^{i} \over i!}$ one immediately obtains using x = −1:

$\lim _{n\to \infty }{!n \over n!}={1 \over e}\approx 0.3679\ldots .$ This is the limit of the probability that a randomly selected permutation of a large number of objects is a derangement. The probability converges to this limit extremely quickly as n increases, which is why !n is the nearest integer to n!/e. The above semi-log graph shows that the derangement graph lags the permutation graph by an almost constant value.

## 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 AS, we often wish to know the number of pairs of functions (fg) 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, ..., nr letters Xr it turns out (after a proper use of the inclusion-exclusion formula) that the answer has the form:

$\int _{0}^{\infty }P_{n_{1}}(x)P_{n_{2}}(x)\cdots P_{n_{r}}(x)e^{-x}\,dx,$ for a certain sequence of polynomials Pn, where Pn has degree n. But the above answer for the case r = 2 gives an orthogonality relation, whence the Pn's are the Laguerre polynomials (up to a sign that is easily decided). $\int _{0}^{\infty }(t-1)^{z}e^{-t}dt$ in the complex plane.

In particular, for the classical derangements

$!n=\int _{0}^{\infty }(x-1)^{n}e^{-x}dx.$ ## 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.