# Derangement

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.

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.

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¡.[1][2]

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[3] in 1708; he solved it in 1713, as did Nicholas Bernoulli at about the same time.

## Example

The 9 derangements (from 24 permutations) are highlighted

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,

 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 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

Counting the derangements of a set amounts to what is known as the hat-check problem,[4] in which one considers the number of ways in which n hats (call them h1 through hn) can be returned to n people (P1 through Pn) such that no hat makes it back to its owner.

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

1. Pi 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 Pj besides Pi, the unreceivable hat is hj, while for Pi it is h1).
2. Pi receives h1. In this case the problem reduces to n − 2 people and n − 2 hats.

For each of the n − 1 hats that P1 may receive, the number of ways that P2, … ,Pn 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

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

where !0 = 1 and !1 = 0. The first few values of !n are:

 n !n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 1 0 1 2 9 44 265 1,854 14,833 133,496 1,334,961 14,684,570 176,214,841 2,290,792,932

There are also various other (equivalent) expressions for !n:[5]

${\displaystyle !n=n!\sum _{i=0}^{n}{\frac {(-1)^{i}}{i!}},\quad n\geq 0,}$
${\displaystyle !n=\left\lfloor {\frac {n!}{e}}\right\rceil =\left\lfloor {\frac {n!}{e}}+{\frac {1}{2}}\right\rfloor ,\quad n\geq 1.}$

(where ${\displaystyle \left\lfloor x\right\rceil }$  is the nearest integer function[6] and ${\displaystyle \left\lfloor x\right\rfloor }$  is the floor function)

For any integer n ≥ 1,

${\displaystyle !n={\begin{cases}\lfloor {\frac {n!}{e}}+r_{1}\rfloor ,&n{\text{ is odd}},\quad \ r_{1}\in [0,{\frac {1}{2}}],\\\lfloor {\frac {n!}{e}}+r_{2}\rfloor ,&n{\text{ is even}},\quad r_{2}\in [{\frac {1}{3}},1].\end{cases}}}$

So, for any integer n ≥ 1, and for any real number r ∈ [1/3, 1/2],

${\displaystyle !n=\left\lfloor {\frac {n!}{e}}+r\right\rfloor ,\quad \ n\geq 1,\quad r\in \left[{\frac {1}{3}},{\frac {1}{2}}\right].}$

Therefore, as e = 2.71828…, 1/e ∈ [1/3, 1/2], then [7]

${\displaystyle !n=\left\lfloor {\frac {n!+1}{e}}\right\rfloor ,\quad \ n\geq 1,}$
${\displaystyle !n=\left\lfloor (e+e^{-1})n!\right\rfloor -\lfloor en!\rfloor ,\quad n\geq 2,}$
${\displaystyle !n=n!-\sum _{i=1}^{n}{n \choose i}\cdot {!(n-i)},\quad \ n\geq 1.}$

The following recurrence equality also holds:[8]

${\displaystyle !n=n\left(!(n-1)\right)+(-1)^{n},\quad \ n\geq 1.}$

### Derivation by inclusion–exclusion principle

Another derivation of an (equivalent) formula for the number of derangements of an n-set is as follows. For ${\displaystyle 1\leq k\leq n}$  we define ${\displaystyle S_{k}}$  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 ${\displaystyle (n-i)!}$  permutations. There are ${\displaystyle {n \choose i}}$  such collections, so the inclusion–exclusion principle yields

{\displaystyle {\begin{aligned}|S_{1}\cup \cdots \cup S_{n}|&=\sum _{i}\left|S_{i}\right|-\sum _{i

and since a derangement is a permutation that leaves none of the n objects fixed, we get

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

## Limit of ratio of derangement to permutation as n approaches ∞

From

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

and

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

one immediately obtains using x = −1:

${\displaystyle \lim _{n\to \infty }{!n \over n!}={1 \over e}\approx 0.3679\ldots .}$

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.

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

## 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:

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

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).[9]

${\displaystyle \int _{0}^{\infty }(t-1)^{z}e^{-t}dt}$  in the complex plane.

In particular, for the classical derangements

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

## 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.[10]

## References

1. ^ The name "subfactorial" originates with William Allen Whitworth; see Cajori, Florian (2011), A History of Mathematical Notations: Two Volumes in One, Cosimo, Inc., p. 77, ISBN 9781616405717.
2. ^ Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics (1994), Addison–Wesley, Reading MA. ISBN 0-201-55802-5
3. ^ 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.
4. ^ Scoville, Richard (1966). "The Hat-Check Problem". American Mathematical Monthly. 73 (3): 262–265. doi:10.2307/2315337. JSTOR 2315337.
5. ^ Hassani, M. "Derangements and Applications." J. Integer Seq. 6, No. 03.1.2, 1–8, 2003
6. ^
7. ^
8. ^ See the notes for (sequence A000166 in the OEIS).
9. ^ Even, S.; J. Gillis (1976). "Derangements and Laguerre polynomials". Mathematical Proceedings of the Cambridge Philosophical Society. 79 (1): 135–143. doi:10.1017/S0305004100052154. Retrieved 27 December 2011.
10. ^ Lubiw, Anna (1981), "Some NP-complete problems similar to graph isomorphism", SIAM Journal on Computing, 10 (1): 11–21, doi:10.1137/0210002, MR 0605600. Babai, László (1995), "Automorphism groups, isomorphism, reconstruction", Handbook of combinatorics, Vol. 1, 2 (PDF), Amsterdam: Elsevier, pp. 1447–1540, MR 1373683, A surprising result of Anna Lubiw asserts that the following problem is NP-complete: Does a given permutation group have a fixed-point-free element?.