The matching Problem 10/9

In class I showed that the probability that a random permutation had at least one fixed point could be computed using the Inclusion-Exclusion Principle for n events:

$$P(E_1\cup E_2 \cup \cdots E_n)= \sum P(E_i)-\sum_{i_1<i_2}P(E... ... +\ldots +(-1)^{(n+1)}P(E_{i_1}\cap E_{i_2}\cap..\cap E_{i_n})$$

taking E_i the vent that the ith point is a fixed point. We can see that $P(E_i)=\frac{(n-1)!}{n!}$and that

$$P(E_i \cap E_j)=\frac{(n-2)!}{n!} \qquad P(E_i \cap E_j \cap E_k) =\frac{(n-3)!}{n!}$$

As there are $\binom{n}{2}$ pairs i<j and $\binom{n}{3}$ triples i<j<k , the formula gives:

$$P(\mbox{at least one match}) =1- \frac{1}{2!}+ \frac{1}{3!} - \frac{1}{4!} + \frac{(-1)^{n-1}}{n!}\equiv 1-\frac{1}{e}$$

Note that this does not depend on n.