2 the act of disturbing the mind or body; "his carelessness could have caused an ecological upset"; "she was unprepared for this sudden overthrow of their normal way of living" [syn: upset, overthrow]
- The property of being deranged
In combinatorial mathematics, a derangement is a permutation in which none of the elements of the set appear in their original positions. 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.
Suppose that a professor has graded 4 tests for 4 students. The first student, student A, received an "A" on the test, student B received a "B", and so on. However, the professor mixed up the tests when handing them back, and now none of the students has the correct test. How many ways could the professor have mixed them all up in this way? Out of 24 possible permutations for handing back the tests, there are only 9 derangements:
- BADC, BCDA, BDAC,
- CADB, CDAB, CDBA,
- DABC, DCAB, DCBA.
- CADB, CDAB, CDBA,
In every other permutation of this 4-member set, at least one student gets the right test.
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 φn is any derangement of the natural numbers , then for some k in , φn(n) = k. Then if we let (k, n) be the permutation of which swaps k and n, and we let φn − 1 be the composition ((k, n) o φn); then φn−1(n) = n, and either:
- φn − 1(k) ≠ k, so φn − 1 is a derangement of , or
- φn−1(k) = k, and for all x ≠ k, φn−1(x) ≠ x.
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 dn as the number of derangements of n letters, and we define d0 = 1, d1 = 0; then dn satisfies the recurrence:
- d_n = (n - 1) (d_ + d_)\,
- d_n = n d_ + (-1)^ , \quad n\geq 1
Notice that this same recurrence formula also works for factorials with different starting values. That is 0! = 1, 1! = 1 and
- n! = (n - 1) ((n-1)! + (n-2)!)\,
which is helpful in proving the limit relationship with e below.
Also, the following formulas are known :
- d_n = n! \sum_^n \frac
- d_n = \left\lfloor\frac+\frac\right\rfloor = \left(\text\frac\right), \quad n\geq 1
- d_n = \left\lfloor(e+e^)n!\right\rfloor-\lfloor en!\rfloor , \quad n\geq 2.
Starting with n = 0, the numbers of derangements, dn, are:
- 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961, 14684570, 176214841, 2290792932, ... .
Limit as n approaches ∞
Using this recurrence, it can be shown that, in the limit,
- \lim_ = \approx 0.3679\dots.
This is the limit of the probability pn = dn/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)).
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 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_ (x) P_(x)\cdots P_(x) e^\, 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 Pns are the Laguerre polynomials (up to a sign that is easily decided).
- 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.
derangement in German: Derangement
derangement in Italian: Dismutazione (matematica)
derangement in Japanese: 完全順列
derangement in Russian: Беспорядок (перестановка)
derangement in Slovak: Dismutácia (matematika)
derangement in Chinese: 亂序
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