A combinatorial proof for an identity proceeds as follows: State a counting question. Then, answer the question in two ways: One answer must correspond to the left-hand side (LHS) of the identity. The other answer must correspond to the right-hand side (RHS). Conclude that the LHS is equal to the RHS.
Using this proof method, give a combinatorial proof of the identity:
$k \choose 2$ + $k \choose k-2$ + $k^2$ = $2k \choose 2$, where $k\geq2$
I understand you have to essentially split the $2k \choose 2$. I could treat it as $m +n \choose 2$ where m and n are equivalent and equal to k. And I get where the $k \choose 2$ and $k \choose k-2$ come from but I have no idea how the $k^2$ comes into play. and why thats involved?
Noting that $\binom{k}{k-2}=\binom{k}{2}$ (choosing $2$ out of $k$ is the same as choosing the $k-2$ in the complement within the $k$), we can see the right as the number of ways to pick $2$ persons out of $2k$ persons, $k$ of which are men and $k$ are women.
The left had side corresponds then to the mixed groups (1 man out of the $k$, 1 woman out of the $k$, so $k^2$ of those) and the all men groups $\binom{k}{2}$ and all women groups (also $\binom{k}{2} = \binom{k}{k-2}$). Together these combine also to all such choices of $2$ out of the $2k$).