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

(Solution 2)
(Solution 2)
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The expected value is also <math>\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}</math>.
 
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>
  
 
{{IMO box|before=First question|num-a=2|year=1987}}
 
{{IMO box|before=First question|num-a=2|year=1987}}
  
 
[[Category:Olympiad Combinatorics Problems]]
 
[[Category:Olympiad Combinatorics Problems]]

Revision as of 13:22, 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 \[\sum_{k=0}^{n} k \cdot p_n (k) = n!.\]

1987 IMO (Problems) • Resources
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