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.

For any size-$3$ subset of $\{1,2,\dots, 2n\}$, trade each element $a$ for $2n+1-a$. This is an involution; that just means that if you do it again you get back to the same thing. Therefore, we can simply pair up the subsets according to this rule.
We might be worried that something gets paired up with itself; this of course would ruin the whole endeavor. But notice that each element in $\{n+1,n+2,\dots, 2n\}$ is sent to an element in $\{1,2,\dots, n\}$ and the opposite is also true. Since there must be an odd number of elements in the subsets we are counting, there must be an odd number of their elements in the bottom half or the top half, but not both. After performing the rule above, which half has the odd number switches, so the set rule could not possibly send any set to itself.
(This proof has the advantage of generalizing very easily to show that $\binom nk$ is even whenever $n$ is even and $k$ is odd. It's the same as Brian's, although he found a more imaginative scenario.)
