I was reading up on Singmaster's conjecture on repeated binomial coefficiencts and I read that $$\binom{15}{5} = \binom{14}{6}$$
Sure, it's possible to prove it non-combinatorically:
$$\begin{align}\binom{15}{5} &= \frac{15!}{5!10!} \\&= \frac{6\cdot15!}{10\cdot9\cdot6!8!} \\&= \frac{14!}{6!8!} \\&= \binom{14}{6}\end{align}$$
However, I was thinking: is it possible to come up with a combinatorical argument that proves the equality of the two?