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

## 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.

### Solution 4

Consider a bipartite graph $G$ with bipartition $\{A,B\}$. The vertices in $A$ are the $(r+1)$-element subsets of $\{0, \dots , n\}$, and the vertices in $B$ are the $r$-element subsets of $\{1, \dots , n\}$, and we draw an edge $\overline{ab}$ iff the subset $b \in B$ may be obtained from $a \in A$ by deleting the smallest element in $a$.

Note that

$|A|=\binom{n+1}{r+1}, |B|=\binom{n}{r}, |E(G)|=\binom{n+1}{r+1}=\frac{n+1}{r+1} \binom{n}{r}.$

The degree of a vertex in $B$ is the value of the least element of its corresponding subset. Hence $$F(n,r)=\frac{1}{\binom{n}{r}} \sum_{v \in B} \deg (v)= \frac{n+1}{r+1}.$$

### Solution 5

We will count how many times each element of the set $\{1, 2, \ldots, n\}$ appear as the smallest element in a set, which will lead us to the result.

To count how many times $1$ appears as the smallest element of a subset, we need to choose an $r-1$ element subset from the remaining $n - 1$ numbers, as all of them are greater than $1.$ There are simply $\dbinom{n - 1}{r - 1}$ ways to do this.

Similarly, to count how many times $2$ appears as the smallest element, we need to choose an $r - 1$ element subset from the remaining $n - 2$ numbers greater than $2.$ There are $\dbinom{n - 2}{r - 1}$ ways to do this.

As there are $\dbinom{n}{r}$ subsets, and thus smallest elements, in total, it now becomes clear that we need to evaluate the sum $$\frac{\binom{n - 1}{r - 1} + 2\binom{n - 2}{r - 1} + \cdots + (n - r + 1)\binom{r - 1}{r - 1}}{\binom{n}{r}}.$$ We can rearrange this sum as $$\frac{\left(\binom{n - 1}{r - 1} + \binom{n - 2}{r - 1} + \cdots + \binom{r - 1}{r - 1}\right)+\left(\binom{n - 2}{r - 1} + \binom{n - 3}{r - 1} + \cdots + \binom{r - 1}{r - 1}\right)+\cdots+ \binom{r-1}{r-1}}{\binom{n}{r}}.$$ Using the Hockey Stick Identity on each of the smaller sums on the numerator gives us $$\frac{\binom{n}{r} + \binom{n - 1}{r} + \binom{n-2}{r} + \cdots + \binom{r}{r}}{\binom{n}{r}}.$$ Using Hockey stick again gives us $$\frac{\binom{n + 1}{r + 1}}{\binom{n}{r}},$$ which simplifies to $$\frac{\frac{(n + 1)!}{(r + 1)!(n-r)!}}{\frac{n!}{r!(n-r)!}} = \left(\frac{(n + 1)!}{(r + 1)!(n - r)!}\right) \left(\frac{r!(n-r)!}{n!}\right) = \frac{n + 1}{r + 1},$$ as desired.

Solution by Ilikeapos