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Given a random permutation $\sigma \in S_n$ from $[n] \to [n]$ in a uniform probability space, what is the probability that $\sigma $ has exactly $k$ fixed points for a given $k$ between $1$ and $n$?

In other words: what is the probability that $\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i $ for $\ i\in \{1,...,k\}$ and for every $y \notin \{x_1 , ... , x_k\}$ we get $\sigma(y) \neq y$.

I saw that $\lim_{n \to \infty } prob(A_0) = e^{-1}$ using Inclusion–exclusion principle and i belive that for a given k : $\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$ but I am not sure how to show it.

*$A_k$ stands for the event "k".

ned grekerzberg
  • 977

3 Answers3

18

There are $\binom{n}{k}$ possibilities for the $k$ fixed points.

If they are established then there are $!(n-k)$ derangements for the other points.

That gives $\binom{n}{k}\left[!(n-k)\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.

Also we have the formula: $$!(n-k)=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$ and we end up with probability: $$\frac1{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$

drhab
  • 151,093
6
  1. Number of dearrangements of $k$-elements set is $$!k = k!\sum\limits_{m=0}^{k}\frac{(-1)^m}{m!}$$
  2. In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\binom{n}{k}$ ways

Thus $$A_k^n = \binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$ And probability is $$P(A_k^n) = \frac{\binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{\frac{n!}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{1}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$

Jaroslaw Matlak
  • 4,895
6

By way of enrichment here is an alternate formulation using combinatorial classes. The class of permutations with fixed points marked is

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$

This gives the generating function $$G(z, u) = \exp\left(uz + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \frac{z^5}{5} + \cdots\right)$$

which is

$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right) = \frac{\exp(uz-z)}{1-z}.$$

Now for $k$ fixed points we get

$$[u^k] \frac{\exp(uz-z)}{1-z} = [u^k] \frac{\exp(uz)\exp(-z)}{1-z} = \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$

This is the EGF of permutations having $k$ fixed points. We extract the count by computing (the factor $n!$ is canceled because we require the average)

$$[z^n] \frac{z^k}{k!} \frac{\exp(-z)}{1-z} = \frac{1}{k!} [z^{n-k}] \frac{\exp(-z)}{1-z}.$$

We find

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{k!} \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}.}$$

We can identify this as choosing the $k$ fixed points and combining them with a derangement of the rest:

$$\frac{1}{n!} {n\choose k} (n-k)! \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}$$

which is the combinatorial class

$$\textsc{SET}_{=k}(\mathcal{Z}) \times \textsc{SET}(\textsc{CYC}_{\ge 2}(\mathcal{Z})).$$

Marko Riedel
  • 61,317