Added to Favorites

Related Searches

Definitions

Nearby Words

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.
## Example

## Counting derangements

## Limit as n approaches ∞

## Generalizations

## References

## External links

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.

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 φ_{n} is any derangement of the natural numbers { 1, ..., n }, then for some k in { 1, ..., n − 1 }, φ_{n}(n) = k. Then if we let (k, n) be the permutation of { 1, ..., n } which swaps k and n, and we let φ_{n − 1} be the composition ((k, n) o φ_{n}); then φ_{n−1}(n) = n, and either:

- φ
_{n − 1}(k) ≠ k, so φ_{n − 1}is a derangement of { 1, ..., n − 1 }, or - φ
_{n−1}(k) = k, and for all x ≠ k, φ_{n−1}(x) ≠ x.

In the latter case, φ_{n − 1} is then a derangement of the set { 1, ..., n − 1 } excluding k; i.e., the composition φ_{n−2} = ((k,n − 1) o φ_{n − 1} o (k,n−1)) is a derangement of { 1, ..., n − 2 }.

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 d_{n} as the number of derangements of n letters, and we define d_{0} = 1, d_{1} = 0; then d_{n} satisfies the recurrence:

- $d\_n\; =\; (n\; -\; 1)\; (d\_\{n-1\}\; +\; d\_\{n-2\}),$

and also

- $d\_n\; =\; n\; d\_\{n-1\}\; +\; (-1)^\{n\}\; ,\; quad\; ngeq\; 1$

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 :

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

- $d\_n\; =\; leftlfloorfrac\{n!\}\{e\}+frac\{1\}\{2\}rightrfloor\; =\; left(text\{nearest\; integer\; to\; \}frac\{n!\}\{e\}right),\; quad\; ngeq\; 1$

- $d\_n\; =\; leftlfloor(e+e^\{-1\})n!rightrfloor-lfloor\; en!rfloor\; ,\; quad\; ngeq\; 2.$

Starting with n = 0, the numbers of derangements, d_{n}, are:

- 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961, 14684570, 176214841, 2290792932, ... .

These numbers are also called subfactorial or rencontres numbers.

Using this recurrence, it can be shown that, in the limit,

- $lim\_\{ntoinfty\}\; \{d\_n\; over\; n!\}\; =\; \{1\; over\; e\}\; approx\; 0.3679dots.$

This is the limit of the probability p_{n} = d_{n}/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 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:

- $int\_0^infty\; P\_\{n\_1\}\; (x)\; P\_\{n\_2\}(x)cdots\; P\_\{n\_r\}(x)\; e^\{-x\},\; dx,$

for a certain sequence of polynomials 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).

- 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.

*

*

*

*

*

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Saturday September 20, 2008 at 01:11:41 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Saturday September 20, 2008 at 01:11:41 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2015 Dictionary.com, LLC. All rights reserved.