1981 IMO Problems/Problem 2

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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:

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

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