If $n\ge2$, Prove $\binom{2n}{3}$ is even. Any help would be appreciated. I can see it's true from pascal's triangle, and I've tried messing around with pascal's identity and the binomial theorem to prove it, but I'm just not making any headway.

In the spirit of various other answers, let $S=\{\color{red}1,\color{blue}1,\color{red}2,\color{blue}2\dots, \color{red}n,\color{blue}n\}$. If $T$ is a $3$-element subset of $S$, one color predominates among its elements, and the set $T'$, obtained by changing the color of every number in $T$, is a different $3$-element subset of $S$. (The other color predominates.) The $3$-element subsets of $S$ can thus be collected into disjoint pairs, each subset paired with its “color opposite,” so there must be an even number of these subsets.
