Mathematical Reflections 2012, Issue 5 - Problem S243

Problem:
A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.

Proposed by Iurie Boreico.

Solution:
Let $m$ be the number of boys and $n$ be the number of girls. There are $\displaystyle{m \choose 2}$ possibile pairs of boys and $\displaystyle{n \choose 2}$ possible pairs of girls. Suppose that $\displaystyle{m \choose 2} > \displaystyle{n \choose 2}$. Then, there are two distinct pairs of boys which dance with a pair of girls, but this means that there are at least three boys which dance with two girls, contradiction. With the same reasoning we can prove that it can't be $\displaystyle{m \choose 2} < \displaystyle{n \choose 2}$, so $\displaystyle{m \choose 2} = \displaystyle{n \choose 2}$, i.e. $m=n$.