Rencontres numbers

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In combinatorial mathematics, 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 Dnk 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 D7, 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, and there are D7, 2 = 924 possibilities once more, now, that 2 previous couples meet again just by chance.

Numerical values[edit]

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

_n\!\!\diagdown\!\!^k 0 1 2 3 4 5 6 7
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


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} = \left[{n! \over e}\right]

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, we have

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 Dn,0/(n!) are generated by the power series ez/(1 − z); accordingly, an explicit formula for Dnm 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} = {n \choose m} D_{n-m, 0} \; \; \mbox{ and } \; \;
\frac{D_{n, m}}{n!} \approx \frac{e^{-1}}{m!}

for n large, m fixed.

Probability distribution[edit]

The sum of the entries in each row is the whole 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

{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.

Limiting probability distribution[edit]

As the size of the permuted set grows, we get

\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.


  1. ^ Jim Pitman, "Some Probabilistic Aspects of Set Partitions", American Mathematical Monthly, volume 104, number 3, March 1997, pages 201–209.