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Difference between revisions of "1981 IMO Problems/Problem 2"

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== Problem ==
 
== Problem ==
  
−
Let <math> \displaystyle 1 \le r \le n </math> and consider all subsets of <math> \displaystyle r </math> elements of the set <math> \{ 1, 2, \ldots , n \} </math>.  Each of these subsets has a smallest member.  Let <math> \displaystyle F(n,r) </math> denote the arithmetic mean of these smallest numbers; prove that
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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
  
 
<center>
 
<center>
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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.
  
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Consider a binary string of length <math> \displaystyle n+1 </math> which contains <math> \displaystyle r+1 </math> 1s.  For some value of <math>k</math> between 1 and <math>\displaystyle n</math>, inclusive, we say that the second 1 will occur in the <math> \displaystyle (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.
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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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<center>
 
<center>
 
<math>
 
<math>
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\begin{matrix} \displaystyle \sum_{k=1}^{n}k{n-k \choose r-1 } &=& \displaystyle \sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\
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\begin{matrix}\sum_{k=1}^{n}k{n-k \choose r-1 } &=&\sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\
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&=& \displaystyle  \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\
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&=& \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\
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&=& \displaystyle \sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\
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&=&\sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\
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&=& \displaystyle {n+1 \choose r+1}
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&=&{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.
  
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We define <math>\displaystyle F(k, k-1)</math> to be zero (the empty sum).
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We define <math>F(k, k-1)</math> to be zero (the empty sum).
  
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We consider <math> \displaystyle r</math> to be fixed.  The assertion obviously holds for <math>\displaystyle r = n</math>.  We now assume the problem to hold for values of <math>\displaystyle n</math> less than or equal to <math> \displaystyle k</math>.  By considering subsets containing <math>\displaystyle k+1</math> and not containing <math>\displaystyle k+1</math>, respectively, we conclude that
+
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
  
 
<center>
 
<center>
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{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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{{IMO box|num-b=1|num-a=3|year=1981}}
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* [[1981 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366639#p366639 Discussion on AoPS/Mathlinks]
 
−
 
 
  
 
[[Category:Olympiad Combinatorics Problems]]
 
[[Category:Olympiad Combinatorics Problems]]

Revision as of 19:45, 25 October 2007

Problem

Let $1 \le r \le n$ and consider all subsets of $r$ elements of the set $\{ 1, 2, \ldots , n \}$. Each of these subsets has a smallest member. Let $F(n,r)$ denote the arithmetic mean of these smallest numbers; prove that

$F(n,r) = \frac{n+1}{r+1}.$

Solutions

Solution 1

Clearly, the sum of the desired least elements is $\sum_{k=1}^{n} k {n-k \choose r-1} = \sum_{k=1}^{n} {k \choose 1}{n-k \choose r-1}$, which we claim to be equal to ${n+1 \choose r+1}$ by virtue of the following argument.

Consider a binary string of length $n+1$ which contains $r+1$ 1s. For some value of $k$ between 1 and $n$, inclusive, we say that the second 1 will occur in the $(k+1)$th place. Clearly, there are ${k \choose 1}$ ways to arrange the bits coming before the second 1, and ${n-k \choose r-1}$ ways to arrange the bits after the second 1. Our identity follows.

Since the sum of the least elements of the sets is ${n+1 \choose r+1}$, the mean of the least elements is $\frac{{n+1 \choose r+1}}{{n \choose r}} = \frac{n+1}{r+1}$, Q.E.D.

Solution 2

We proceed as in the previous solution, but we prove our identity using the following manipulations:

$\begin{matrix}\sum_{k=1}^{n}k{n-k \choose r-1 } &=&\sum_{k=1}^{n-(r-1)}{k}{n-k \choose r-1}\\ &=& \sum_{i=1}^{n-(r-1)}\sum_{k=i}^{n-(r-1)}{n-k \choose r-1}\\ &=&\sum_{i=1}^{n-(r-1)}{n-i+1 \choose r}\\ &=&{n+1 \choose r+1} \end{matrix}$

Q.E.D.

Solution 3

We proceed by strong induction.

We define $F(k, k-1)$ to be zero (the empty sum).

We consider $r$ to be fixed. The assertion obviously holds for $r = n$. We now assume the problem to hold for values of $n$ less than or equal to $k$. By considering subsets containing $k+1$ and not containing $k+1$, respectively, we conclude that

$F(k+1, r) = \frac{{k \choose r-1}F(k,r-1) + {k \choose r}F(k,r)}{{k+1 \choose r}} = 1 + \frac{k-r+1}{r+1} = \frac{k+2}{r+1}$.

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
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