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 opposite-sex couples, where, after tea-break the participants are told to randomly find an opposite-sex partner to continue, then once more there are D_7, 2 = 924 possibilities that exactly 2 previous couples meet again by chance.

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

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

${\displaystyle D_{0,0}=1,\!}$

${\displaystyle D_{1,0}=0,\!}$

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

for non-negative n. It turns out that

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

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 ${\displaystyle k\geq 0}$, we have

${\displaystyle D_{n,k}={n \choose k}\cdot D_{n-k,0}.}$

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:

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

This immediately implies that

${\displaystyle D_{n,m}={n \choose m}D_{n-m,0}\;\;{\mbox{ and }}\;\;{\frac {D_{n,m}}{n!}}\approx {\frac {e^{-1}}{m!}}}$

for n large, m fixed.

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

${\displaystyle {D_{n,k} \over n!}.}$

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.^[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.

As the size of the permuted set grows, we get

${\displaystyle \lim _{n\to \infty }{D_{n,k} \over n!}={e^{-1} \over k!}.}$

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.

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

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