Difference between revisions of "1987 IMO Problems/Problem 1"
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as desired. | as desired. | ||
| + | ==Solution 2== | ||
| + | The probability of any number <math>i</math> where <math>1\le i\le n</math> being a fixed point is <math>\frac{1}{n}</math>. Thus, the expected value of the number of fixed points is <math>n\times \frac{1}{n}=1</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 \[\sum_{k=0}^{n} k \cdot p_n (k) = n!]\. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 13:21, 21 July 2012
Problem
Let 
be the number of permutations of the set 
, which have exactly 
fixed points. Prove that
.
(Remark: A permutation 
of a set 
is a one-to-one mapping of onto itself. An element 
in is called a fixed point of the permutation if 
.)
Solution
The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element of the set, there are 
permutations which have as a fixed point. Therefore
,
as desired.
Solution 2
The probability of any number where 
being a fixed point is 
. Thus, the expected value of the number of fixed points is 
.
The expected value is also 
.
Thus, ![\[\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}=1\]](http://latex.artofproblemsolving.com/c/3/b/c3b7d1bac792f986ac46ed91b5bf43f8ced623d9.png)
, 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 | ||
