IMO 1995 Problem 6

Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?

=============================

Solution:
Consider set $P=\{0,1,\ldots,p-1\}$ and its two equal-sized subsets $B$ and $C$ such that their sums are identical modulo $p$. We claim that there's bijection between $(B,C)$ and $A$ as follows. For any $x\in B\cap C$, we add $x$ to $A$; for any $x\in B-C$, we add both $x$ and $x+p$ to $A$; for any $x\notin B\cup C$, we add $x+p$ to $A$. Clearly $|A|=p$ and sum up to $0$ modulo $p$. So we count $(B,C)$ below.

Let $k\notin\{0, p\}$ be the size of $B$ and $C$, and $K$ be the set of $k$-element subsets. Because $p$ is prime, $K$ is equally divided into $p$ disjoint partitions by subset sum modulo $p$. Each subset has size $\frac{1}{p}\binom{p}{k}$ and contributes $\frac{1}{p^2}\binom{p}{k}^2$ to count of ordered pair $(B,C)$. So each $k\notin\{0,p\}$ contributes $\frac{1}{p}\binom{p}{k}^2$, and the total sum is

$$
\frac{1}{p}\sum_{k=0}^{p}\binom{p}{k}^2-\frac{2}{p}+2=\frac{1}{p}\binom{2p}{p}-\frac{2}{p}+2
$$