1987 IMO Problems/Problem 1

Revision as of 13:22, 21 July 2012 by Numbertheorist17 (talk | contribs) (Solution 2)


Let $p_n (k)$ be the number of permutations of the set $\{ 1, \ldots , n \} , \; n \ge 1$, which have exactly $k$ fixed points. Prove that

$\sum_{k=0}^{n} k \cdot p_n (k) = n!$.

(Remark: A permutation $f$ of a set $S$ is a one-to-one mapping of $S$ onto itself. An element $i$ in $S$ is called a fixed point of the permutation $f$ if $f(i) = i$.)


The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element $i$ of the set, there are $(n-1)!$ permutations which have $i$ as a fixed point. Therefore

$\sum_{k=0}^{n} k \cdot p_n (k) = n!$,

as desired.

Solution 2

The probability of any number $i$ where $1\le i\le n$ being a fixed point is $\frac{1}{n}$. Thus, the expected value of the number of fixed points is $n\times \frac{1}{n}=1$.

The expected value is also $\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}$.

Thus, \[\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1\] or \[\sum_{k=0}^{n} k \cdot p_n (k) = n!.\]

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