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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:

\begin{displaymath}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})
\end{displaymath}

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

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

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

\begin{displaymath}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}\end{displaymath}

Note that this does not depend on n.



Susan Holmes
1998-12-07