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.
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 (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.
Counting the derangements of a set amounts to what is known as the hat-check problem, in which one considers the number of ways in which "n" hats (call them "h"1 through "hn") can be returned to "n" people ("P"1 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 "P"1 receives "hi" and consider "hi"’s owner: "Pi" receives either "P"1's hat, "h"1, or some other. Accordingly, the problem splits into two possible cases:
For each of the "n" − 1 hats that "P"1 may receive, the number of ways that "P"2, … ,"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
where !0 = 1 and !1 = 0. The first few values of !"n" are:
There are also various other (equivalent) expressions for !"n":
For any integer ,
So, for any integer , and for any real number ,
Therefore, as , , then 
The following recurrence also holds:
Derivation by inclusion–exclusion principle.
Another derivation of an (equivalent) formula for the number of derangements of an "n"-set is as follows. For formula_12 we define formula_13 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 formula_14 permutations. There are formula_15 such collections, so the inclusion–exclusion principle yields
and since a derangement is a permutation that leaves none of the "n" objects fixed, we get
Limit of ratio of derangement to permutation as "n" approaches ∞.
From
and
one immediately obtains using "x" = −1:
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 "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:
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 "n"1 letters "X"1, "n"2 letters "X"2, ..., "n""r" letters "X""r" it turns out (after a proper use of the inclusion-exclusion formula) that the answer has the form:
for a certain sequence of polynomials "P""n", where "P""n" has degree "n". But the above answer for the case "r" = 2 gives an orthogonality relation, whence the "P""n"'s are the Laguerre polynomials (up to a sign that is easily decided).
In particular, for the classical derangements
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.