IMO 1995- Problem 6

Problem: Let $p$ be an odd prime. Find the number of $p$- element subsets $A$ of $\left\{1,2,...,2p\right\}$ such that the sum if all elements of $A$ is divisible by $p$.
Solution: For any $p$-element subset $A$ of $\left\{1,2,...,2p\right\}$, denote by $s(A)$ the sum of the elements of $A$. Of the $C_{2p}^{p}$ such subsets, $B=\left\{1,2,...,p\right\}$ and $C=\left\{p+1,p+2,...,2p\right\}$ satisfy $s(B)\equiv s(C)\equiv 0(\text{ mod }p)$. For $A\ne B,C$, we have $A\cap B\ne \emptyset \ne A\cap C$. Partition the $C_{2p}^{p}-2p-$ element subsets other than $b$ and $C$ into groups of size $p$ as follows. Two subsets $A$ and $A'$ are in the same group if and only if $A'\cap C=A\cap C$ and $A'\cap B$ is a cyclic permutation of $A\cap B$ within $B$. Suppose $A\cap B$ has $n$ elements, $0\le n<p$. For some $m$ such that $0<m<p$,
$A'\cap B=\left\{x+m:x\in A\cap B,x+m\le p\right\}\cup \left\{x+m-p:x\in A\cap B,x\le p<x+m\right\}$.
Hence $s(A')-s(A)\equiv mn(\text{ mod }p)$, but $mn$ is not divisible by $p$. It follows that exactly one subset $A$ in each group satisfies $s(A)\equiv 0(\text{ mod }p)$, and the total number of such subsets is $p^-1(C_{2p}^{p}-2)+2$.
Second solution: Let $\omega$ be a primitive $pth$ root of unity. Then
$a_0+a_1\omega+...+a_{p-1}\omega^{p-1}=0$ iff $a_0=a_1=...=a_{p-1}$
Also $\prod\limits_{k=1}^{2p}(x-\omega^k)=(x^{p}-1)^2=2^{2p}-2x^{p}+1$.
If $t(\omega)=\sum a_i\omega^{i}$, then $a_i$ is the number of combinations with $i_1+...+i_p\equiv i(\text{ mod }p)$.
Therefore :$(a_0-2)+a_1\omega+...+a_{p-1}^{p-1}\equiv 0$
and $a_0-2=a_1=...=a_p$. So $|S|=\frac{C_{2p}^{p}-2}{p}+2$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf