Difference between revisions of "1981 IMO Problems/Problem 2"
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== Problem == | == Problem == | ||
− | Let <math> | + | Let <math>1 \le r \le n </math> and consider all subsets of <math>r </math> elements of the set <math> \{ 1, 2, \ldots , n \} </math>. Each of these subsets has a smallest member. Let <math>F(n,r) </math> denote the arithmetic mean of these smallest numbers; prove that |
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Clearly, the sum of the desired least elements is <math> \sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1} </math>, which we claim to be equal to <math>{n+1 \choose r+1} </math> by virtue of the following argument. | Clearly, the sum of the desired least elements is <math> \sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1} </math>, which we claim to be equal to <math>{n+1 \choose r+1} </math> by virtue of the following argument. | ||
− | Consider a binary string of length <math> | + | Consider a binary string of length <math>n+1 </math> which contains <math>r+1 </math> 1s. For some value of <math>k</math> between 1 and <math>n</math>, inclusive, we say that the second 1 will occur in the <math>(k+1) </math>th place. Clearly, there are <math> {k \choose 1} </math> ways to arrange the bits coming before the second 1, and <math>{n-k \choose r-1} </math> ways to arrange the bits after the second 1. Our identity follows. |
Since the sum of the least elements of the sets is <math>{n+1 \choose r+1}</math>, the mean of the least elements is <math>\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}</math>, Q.E.D. | Since the sum of the least elements of the sets is <math>{n+1 \choose r+1}</math>, the mean of the least elements is <math>\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}</math>, Q.E.D. | ||
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<math> | <math> | ||
− | \begin{matrix} | + | \begin{matrix}\sum_{k=1}^{n}k{n-k \choose r-1 } &=&\sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\ |
− | &=& | + | &=& \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\ |
− | &=& | + | &=&\sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\ |
− | &=& | + | &=&{n+1 \choose r+1} |
\end{matrix} | \end{matrix} | ||
</math> | </math> | ||
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We proceed by strong induction. | We proceed by strong induction. | ||
− | We define <math> | + | We define <math>F(k, k-1)</math> to be zero (the empty sum). |
− | We consider <math> | + | We consider <math>r</math> to be fixed. The assertion obviously holds for <math>r = n</math>. We now assume the problem to hold for values of <math>n</math> less than or equal to <math>k</math>. By considering subsets containing <math>k+1</math> and not containing <math>k+1</math>, respectively, we conclude that |
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{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | {{IMO box|num-b=1|num-a=3|year=1981}} |
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[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Revision as of 19:45, 25 October 2007
Problem
Let and consider all subsets of elements of the set . Each of these subsets has a smallest member. Let denote the arithmetic mean of these smallest numbers; prove that
Solutions
Solution 1
Clearly, the sum of the desired least elements is , which we claim to be equal to by virtue of the following argument.
Consider a binary string of length which contains 1s. For some value of between 1 and , inclusive, we say that the second 1 will occur in the th place. Clearly, there are ways to arrange the bits coming before the second 1, and ways to arrange the bits after the second 1. Our identity follows.
Since the sum of the least elements of the sets is , the mean of the least elements is , Q.E.D.
Solution 2
We proceed as in the previous solution, but we prove our identity using the following manipulations:
Q.E.D.
Solution 3
We proceed by strong induction.
We define to be zero (the empty sum).
We consider to be fixed. The assertion obviously holds for . We now assume the problem to hold for values of less than or equal to . By considering subsets containing and not containing , respectively, we conclude that
.
This completes our induction, Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1981 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |