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.

Table of values
${\displaystyle n}$	Permutations, ${\displaystyle n!}$	Derangements, ${\displaystyle !n}$	${\displaystyle {\frac {!n}{n!}}}$
0	1 =1×10^0	1 =1×10^0	= 1
1	1 =1×10^0	0	= 0
2	2 =2×10^0	1 =1×10^0	= 0.5
3	6 =6×10^0	2 =2×10^0	≈0.33333 33333
4	24 =2.4×10^1	9 =9×10^0	= 0.375
5	120 =1.20×10^2	44 =4.4×10^1	≈0.36666 66667
6	720 =7.20×10^2	265 =2.65×10^2	≈0.36805 55556
7	5 040 =5.04×10^3	1 854 ≈1.85×10^3	≈0.36785 71429
8	40 320 ≈4.03×10^4	14 833 ≈1.48×10^4	≈0.36788 19444
9	362 880 ≈3.63×10^5	133 496 ≈1.33×10^5	≈0.36787 91887
10	3 628 800 ≈3.63×10^6	1 334 961 ≈1.33×10^6	≈0.36787 94643
11	39 916 800 ≈3.99×10^7	14 684 570 ≈1.47×10^7	≈0.36787 94392
12	479 001 600 ≈4.79×10^8	176 214 841 ≈1.76×10^8	≈0.36787 94413
13	6 227 020 800 ≈6.23×10^9	2 290 792 932 ≈2.29×10^9	≈0.36787 94412
14	87 178 291 200 ≈8.72×10^10	32 071 101 049 ≈3.21×10^10	≈0.36787 94412
15	1 307 674 368 000 ≈1.31×10^12	481 066 515 734 ≈4.81×10^11	≈0.36787 94412
16	20 922 789 888 000 ≈2.09×10^13	7 697 064 251 745 ≈7.70×10^12	≈0.36787 94412
17	355 687 428 096 000 ≈3.56×10^14	130 850 092 279 664 ≈1.31×10^14	≈0.36787 94412
18	6 402 373 705 728 000 ≈6.40×10^15	2 355 301 661 033 953 ≈2.36×10^15	≈0.36787 94412
19	121 645 100 408 832 000 ≈1.22×10^17	44 750 731 559 645 106 ≈4.48×10^16	≈0.36787 94412
20	2 432 902 008 176 640 000 ≈2.43×10^18	895 014 631 192 902 121 ≈8.95×10^17	≈0.36787 94412
21	51 090 942 171 709 440 000 ≈5.11×10^19	18 795 307 255 050 944 540 ≈1.88×10^19	≈0.36787 94412
22	1 124 000 727 777 607 680 000 ≈1.12×10^21	413 496 759 611 120 779 881 ≈4.13×10^20	≈0.36787 94412
23	25 852 016 738 884 976 640 000 ≈2.59×10^22	9 510 425 471 055 777 937 262 ≈9.51×10^21	≈0.36787 94412
24	620 448 401 733 239 439 360 000 ≈6.20×10^23	228 250 211 305 338 670 494 289 ≈2.28×10^23	≈0.36787 94412
25	15 511 210 043 330 985 984 000 000 ≈1.55×10^25	5 706 255 282 633 466 762 357 224 ≈5.71×10^24	≈0.36787 94412
26	403 291 461 126 605 635 584 000 000 ≈4.03×10^26	148 362 637 348 470 135 821 287 825 ≈1.48×10^26	≈0.36787 94412
27	10 888 869 450 418 352 160 768 000 000 ≈1.09×10^28	4 005 791 208 408 693 667 174 771 274 ≈4.01×10^27	≈0.36787 94412
28	304 888 344 611 713 860 501 504 000 000 ≈3.05×10^29	112 162 153 835 443 422 680 893 595 673 ≈1.12×10^29	≈0.36787 94412
29	8 841 761 993 739 701 954 543 616 000 000 ≈8.84×10^30	3 252 702 461 227 859 257 745 914 274 516 ≈3.25×10^30	≈0.36787 94412
30	265 252 859 812 191 058 636 308 480 000 000 ≈2.65×10^32	97 581 073 836 835 777 732 377 428 235 481 ≈9.76×10^31	≈0.36787 94412

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, D_n, d_n, 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 h_n) can be returned to n people (P₁ through P_n) 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 h_i and consider h_i’s owner: P_i receives either P₁'s hat, h₁, or some other. Accordingly, the problem splits into two possible cases:

1. P_i 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 P_j besides P_i, the unreceivable hat is h_j, while for P_i it is h₁).
2. P_i 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, … ,P_n 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:

The Number of Derangements of an n-Element Set (sequence A000166 in the OEIS) for Small n

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13
!n	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

${\displaystyle {\begin{aligned}|S_{1}\cup \cdots \cup S_{n}|&=\sum _{i}\left|S_{i}\right|-\sum _{i<j}\left|S_{i}\cap S_{j}\right|+\sum _{i<j<k}\left|S_{i}\cap S_{j}\cap S_{k}\right|+\cdots +(-1)^{n+1}\left|S_{1}\cap \cdots \cap S_{n}\right|\\&={n \choose 1}(n-1)!-{n \choose 2}(n-2)!+{n \choose 3}(n-3)!-\cdots +(-1)^{n+1}{n \choose n}0!\\&=\sum _{i=1}^{n}(-1)^{i+1}{n \choose i}(n-i)!\\&=n!\ \sum _{i=1}^{n}{(-1)^{i+1} \over i!}\end{aligned}}}$

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:

${\displaystyle \lim _{n\to \infty }{!n \over n!}={1 \over e}\approx 0.367879\ldots .}$

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 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 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, ..., n_r letters X_r 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 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).[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. Weisstein, Eric W. "Nearest Integer Function". MathWorld.
7. Weisstein, Eric W. "Subfactorial". MathWorld.
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?.

External links

• Baez, John (2003). "Let's get deranged!" (PDF).
• Bogart, Kenneth P.; Doyle, Peter G. (1985). "Non-sexist solution of the ménage problem".
• Dickau, Robert M. "Derangement diagrams". Mathematical Figures Using Mathematica.
• 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.