Derangement
Table of values | |||
---|---|---|---|
Permutations, | Derangements, | ||
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. 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¡.[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
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
Suppose that there are n people 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 the same number as their 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:
- 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).
- 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:
where !n, known as the subfactorial, represents the number of derangements, with the starting values !0 = 1 and !1 = 0.
Also, the following equalities are known:[4]
where is the nearest integer function and is the floor function.
The following recurrence equality also holds:[5]
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).
To derive a formula for the number of derangements of n objects, one can proceed as follows. For we define 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 permutations. There are 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:
- 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, ..., nr letters Xr it turns out (after a proper use of the inclusion-exclusion formula) that the answer has the form:
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 Pn's are the Laguerre polynomials (up to a sign that is easily decided).[6]
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.[7]
References
- ^ 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.
- ^ Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics (1994), Addison–Wesley, Reading MA. ISBN 0-201-55802-5
- ^ 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.
- ^ Hassani, M. "Derangements and Applications." J. Integer Seq. 6, No. 03.1.2, 1–8, 2003
- ^ See the notes for (sequence A000166 in the OEIS).
- ^ Even, S.; J. Gillis (1976). "Derangements and Laguerre polynomials". Mathematical Proceedings of the Cambridge Philosophical Society. 79 (01): 135–143. doi:10.1017/S0305004100052154. Retrieved 27 December 2011.
- ^ 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.