Derangement

For the psychological condition, see psychosis.

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.

Table of values
$n$	Permutations, $n!$	Derangements, $!n$	${\frac {!n}{n!}}$
0	1 =1×100	1 =1×100	= 1
1	1 =1×100	0	= 0
2	2 =2×100	1 =1×100	= 0.5
3	6 =6×100	2 =2×100	≈0.33333 33333
4	24 =2.4×101	9 =9×100	= 0.375
5	120 =1.20×102	44 =4.4×101	≈0.36666 66667
6	720 =7.20×102	265 =2.65×102	≈0.36805 55556
7	5 040 ≈5.04×103	1 854 ≈1.85×103	≈0.36785 71429
8	40 320 ≈4.03×104	14 833 ≈1.48×104	≈0.36788 19444
9	362 880 ≈3.63×105	133 496 ≈1.33×105	≈0.36787 91887
10	3 628 800 ≈3.63×106	1 334 961 ≈1.33×106	≈0.36787 94643
11	39 916 800 ≈3.99×107	14 684 570 ≈1.47×107	≈0.36787 94392
12	479 001 600 ≈4.79×108	176 214 841 ≈1.76×108	≈0.36787 94413
13	6 227 020 800 ≈6.23×109	2 290 792 932 ≈2.29×109	≈0.36787 94412
14	87 178 291 200 ≈8.72×1010	32 071 101 049 ≈3.21×1010	≈0.36787 94412
15	1 307 674 368 000 ≈1.31×1012	481 066 515 734 ≈4.81×1011	≈0.36787 94412
16	20 922 789 888 000 ≈2.09×1013	7 697 064 251 745 ≈7.70×1012	≈0.36787 94412
17	355 687 428 096 000 ≈3.56×1014	130 850 092 279 664 ≈1.31×1014	≈0.36787 94412
18	6 402 373 705 728 000 ≈6.40×1015	2 355 301 661 033 953 ≈2.36×1015	≈0.36787 94412
19	121 645 100 408 832 000 ≈1.22×1017	44 750 731 559 645 106 ≈4.48×1016	≈0.36787 94412
20	2 432 902 008 176 640 000 ≈2.43×1018	895 014 631 192 902 121 ≈8.95×1017	≈0.36787 94412
21	51 090 942 171 709 440 000 ≈5.11×1019	18 795 307 255 050 944 540 ≈1.88×1019	≈0.36787 94412
22	1 124 000 727 777 607 680 000 ≈1.12×1021	413 496 759 611 120 779 881 ≈4.13×1020	≈0.36787 94412
23	25 852 016 738 884 976 640 000 ≈2.59×1022	9 510 425 471 055 777 937 262 ≈9.51×1021	≈0.36787 94412
24	620 448 401 733 239 439 360 000 ≈6.20×1023	228 250 211 305 338 670 494 289 ≈2.28×1023	≈0.36787 94412
25	15 511 210 043 330 985 984 000 000 ≈1.55×1025	5 706 255 282 633 466 762 357 224 ≈5.71×1024	≈0.36787 94412
26	403 291 461 126 605 635 584 000 000 ≈4.03×1026	148 362 637 348 470 135 821 287 825 ≈1.48×1026	≈0.36787 94412
27	10 888 869 450 418 352 160 768 000 000 ≈1.09×1028	4 005 791 208 408 693 667 174 771 274 ≈4.01×1027	≈0.36787 94412
28	304 888 344 611 713 860 501 504 000 000 ≈3.05×1029	112 162 153 835 443 422 680 893 595 673 ≈1.12×1029	≈0.36787 94412
29	8 841 761 993 739 701 954 543 616 000 000 ≈8.84×1030	3 252 702 461 227 859 257 745 914 274 516 ≈3.25×1030	≈0.36787 94412
30	265 252 859 812 191 058 636 308 480 000 000 ≈2.65×1032	97 581 073 836 835 777 732 377 428 235 481 ≈9.76×1031	≈0.36787 94412

In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position.

The number of derangements of a set of size n, usually written D_n, d_n, or !n, is called the "derangement number" or "de Montmort number". (These numbers are generalized to rencontres numbers.) The subfactorial function (not to be confused with the factorial n!) maps n to !n. No standard notation for subfactorials is agreed upon; n¡ is sometimes used instead of !n.

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

The 9 derangements (from 24 permutations) are highlighted

Suppose that a professor has had 4 of his students – student A, student B, student C, and student D – take a test and wants to let his students 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:

BADC, BCDA, BDAC,

CADB, CDAB, CDBA,

DABC, DCAB, DCBA.

In every other permutation of this 4-member set, at least one student gets his or her own test back.

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 persons 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 same number as his/her 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.

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 formulae 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.

In Microsoft Excel, the following equivalent expression is accurate to 14 digits:

   !n = INT( FACT(n)/EXP(1) +0.5) where n>=1.

$!n=\left\lfloor (e+e^{-1})n!\right\rfloor -\lfloor en!\rfloor ,\quad n\geq 2,$

$!n=n!-\sum _{i=1}^{n}{n \choose i}\cdot !(n-i),$

The following recurrence relationship also holds:

$!n=n[!(n-1)]+(-1)^{n}$

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).

These numbers are also called subfactorial or rencontres numbers.

Perhaps a more well-known method of counting derangements uses the inclusion-exclusion principle.

Limit of ratio of derangement to permutation as n approaches ∞

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

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

This is the limit of the probability p_n = d_n/n! that a randomly selected permutation is a derangement. The probability converges to this limit extremely quickly as n increases, which is why d_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:

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, ..., 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).

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.

External links

• Baez, John (2003). "Let's get deranged!" (PDF).
• Bogart, Kenneth P. and Doyle, Peter G. (1985). "Non-sexist solution of the ménage problem".{{cite web}}: CS1 maint: multiple names: authors list (link)
• Dickau, Robert M. "Derangement diagrams". Figures Using Mathematica. {{cite web}}: External link in |work= (help); Italic or bold markup not allowed in: |work= (help)
• Hassani, Mehdi. "Derangements and Applications". Journal of Integer Sequences (JIS), Volume 6, Issue 1, Article 03.1.2, 2003.
• Weisstein, Eric W. "Derangement". MathWorld–A Wolfram Web Resource.