<s>[INST] 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?
 [/INST] 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)
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