Combinatorial proof of $\binom{nk}{2}=k\binom{n}{2}+n^2\binom{k}{2}$
Solution 1:
I don't think you need as many cases, which saves a little algebra. We have $k$ groups of $n$ dots each, so choosing two of them can be done in $\binom{nk}{2}$ ways.
Alternatively, both are in the same group of $n$ which has $\binom{k}{1} \cdot \binom{n}{2}$ possibilities, or they are in different groups of $n$, which has $\binom{k}{2} \binom{n}{1}^2$ possibilities.
Therefore, $\binom{nk}{2} = k \binom{n}{2} + n^2 \binom{k}{2}$