How many "Two pair" poker hands are there in a standard deck? From the solutions to our midterm, I know the answer of this question to be:
$$\binom{13}{2} \times \binom{4}{2} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$
However, I initially thought it would be:
$$\binom{13}{1} \times \binom{4}{2} \times \binom{12}{1} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$
I think I see that my answer would could KKQQ5 and QQKK5 as two separate hands, but wouldn't the solution read 5QQKK and QQKK5 as separate hands as well?

The difference between yours and the solution is $\binom{13}{2}$ versus $\binom{13}{1} \times \binom{12}{1}$. If you write them out you will see that $2 \binom{13}{2} = \binom{13}{1} \times \binom{12}{1}$, the 2 double counting as you have noticed (that KKQQ5 and QQKK5 are essentially the same). 
(Note that $\binom{13}{2}$, the number of ways to choose $2$ pairs from 13 possible is unordered. Meanwhile, $\binom{13}{1} \times \binom{12}{1}$ involves order which we have to remove)
