Difference between revisions of "1987 IMO Problems/Problem 1"

Revision as of 12:22, 21 July 2012

Problem

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 $f(i) = i$ .)

Solution

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 $\sum_{k=0}^{n} k \cdot p_n (k) = n!$ . Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1987 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

@@ Line 29: / Line 29: @@
 The expected value is also <math>\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}</math>.
-Thus, <cmath>\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1</cmath>, or <cmath>\sum_{k=0}^{n} k \cdot p_n (k) = n!</cmath>.
+Thus, <cmath>\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1</cmath> or <cmath>\sum_{k=0}^{n} k \cdot p_n (k) = n!</cmath>.
 {{alternate solutions}}

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Difference between revisions of "1987 IMO Problems/Problem 1"

Revision as of 12:22, 21 July 2012

Problem

Solution

Solution 2