Rencontres numbers

In combinatorics, the rencontres numbers are a triangular array of integers that enumerate permutations of the set { 1, ..., n } with specified numbers of fixed points: in other words, partial derangements. (Rencontre is French for encounter. By some accounts, the problem is named after a solitaire game.) For n ≥ 0 and 0 ≤ k ≤ n, the rencontres number D_n, k is the number of permutations of { 1, ..., n } that have exactly k fixed points. For example, if seven presents are given to seven different people, but only two are destined to get the right present, there are D_7, 2 = 924 ways this could happen. Another often cited example is that of a dance school with 7 couples, where, after tea-break the participants are told to randomly find a partner to continue, then once more there are D_7, 2 = 924 possibilities that 2 previous couples meet again by chance.

Numerical values

Here is the beginning of this array (sequence A008290 in the OEIS):

k n	0	1	2	3	4	5	6	7	8
0	1
1	0	1
2	1	0	1
3	2	3	0	1
4	9	8	6	0	1
5	44	45	20	10	0	1
6	265	264	135	40	15	0	1
7	1854	1855	924	315	70	21	0	1
8	14833	14832	7420	2464	630	112	28	0	1

ordered by number of moved elements
The usual way (table above) to show the rencontres numbers is in columns corresponding to the number of fixed points $k$. But one can also order them in columns corresponding to the number of moved elements $i=n-k$ (sequence A098825 in the OEIS). Each entry in the table below on the left can be factored into two terms given in the table below on the right: the product of a binomial coefficient (given first in red) and a subfactorial (given second in blue). In this order each column corresponds to one subfactorial: $T(n,i)=(\genfrac{}{}{0ex}{}{n}{i})\cdot !i$
i n  0 1 2 3 4 5 6 7 8 0 1 1 1 0 2 1 0 1 3 1 0 3 2 4 1 0 6 8 9 5 1 0 10 20 45 44 6 1 0 15 40 135 264 265 7 1 0 21 70 315 924 1855 1854 8 1 0 28 112 630 2464 7420 14832 14833	i n  0 1 2 3 4 5 6 7 8 0 1⋅1 1 1⋅1 1⋅0 2 1⋅1 2⋅0 1⋅1 3 1⋅1 3⋅0 3⋅1 1⋅2 4 1⋅1 4⋅0 6⋅1 4⋅2 1⋅9 5 1⋅1 5⋅0 10⋅1 10⋅2 5⋅9 1⋅44 6 1⋅1 6⋅0 15⋅1 20⋅2 15⋅9 6⋅44 1⋅265 7 1⋅1 7⋅0 21⋅1 35⋅2 35⋅9 21⋅44 7⋅265 1⋅1854 8 1⋅1 8⋅0 28⋅1 56⋅2 70⋅9 56⋅44 28⋅265 8⋅1854 1⋅14833

Formulas

The numbers in the k = 0 column enumerate derangements. Thus

$D_{0,0}=1,$

$D_{1,0}=0,$

$D_{n+2,0}=(n+1)(D_{n+1,0}+D_{n,0})$

for non-negative n. It turns out that

$D_{n,0}=\lceil \frac{n!}{e}\rfloor ,$

where the ratio is rounded up for even n and rounded down for odd n. For n ≥ 1, this gives the nearest integer. More generally, for any $k\ge 0$, we have

$D_{n,k}=(\genfrac{}{}{0ex}{}{n}{k})\cdot D_{n-k,0}.$

The proof is easy after one knows how to enumerate derangements: choose the k fixed points out of n; then choose the derangement of the other n − k points. The numbers D_n,0/(n!) are generated by the power series e^−z/(1 − z); accordingly, an explicit formula for D_n, m can be derived as follows:

$D_{n,m}=\frac{n!}{m!}[z^{n-m}]\frac{e^{-z}}{1-z}=\frac{n!}{m!}\sum _{k=0}^{n-m}\frac{(-1{)}^k}{k!}.$

This immediately implies that

$D_{n,m}=(\genfrac{}{}{0ex}{}{n}{m})D_{n-m,0}\text{ and }\frac{D_{n,m}}{n!}\approx \frac{e^{-1}}{m!}$

for n large, m fixed.

Probability distribution

The sum of the entries in each row for the table in "Numerical Values" is the total number of permutations of { 1, ..., n }, and is therefore n!. If one divides all the entries in the nth row by n!, one gets the probability distribution of the number of fixed points of a uniformly distributed random permutation of { 1, ..., n }. The probability that the number of fixed points is k is

$\frac{D_{n,k}}{n!}.$

For n ≥ 1, the expected number of fixed points is 1 (a fact that follows from linearity of expectation). More generally, for i ≤ n, the ith moment of this probability distribution is the ith moment of the Poisson distribution with expected value 1. For i > n, the ith moment is smaller than that of that Poisson distribution. Specifically, for i ≤ n, the ith moment is the ith Bell number, i.e. the number of partitions of a set of size i.

Limiting probability distribution

As the size of the permuted set grows, we get

$\underset{n\to \mathrm{\infty }}{\lim }\frac{D_{n,k}}{n!}=\frac{e^{-1}}{k!}.$

This is just the probability that a Poisson-distributed random variable with expected value 1 is equal to k. In other words, as n grows, the probability distribution of the number of fixed points of a random permutation of a set of size n approaches the Poisson distribution with expected value 1.

See also

• Oberwolfach problem, a different mathematical problem involving the arrangement of diners at tables
• Problème des ménages, a similar problem involving partial derangements

References

• Riordan, John, An Introduction to Combinatorial Analysis, New York, Wiley, 1958, pages 57, 58, and 65.
• Weisstein, Eric W. "Partial Derangements". MathWorld.