Counting problem for finding the probability of two pairs in a five card hand.

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I am having trouble understanding what I have wrong in this stats problem for my hw. I am trying to find the probability of getting two pairs in a 5 card hand from a deck of cards. So first we know that $$P(\text{Two pairs})=\frac{N(\text{Two pairs})}{N(\text{total 5 card hands})}$$ I know the answer to this question is $$\frac{{13 \choose 2}{4 \choose 2}{4 \choose 2}{11 \choose 1}{4 \choose 1}}{{52 \choose 5}},$$ which makes sense because we choose 2 ranks out of the 13 times choose 2 pairs times the last card which will be a different rank. However, I got this answer (which I know is wrong) $$\frac{{13 \choose 1}{4 \choose 2}{12 \choose 1}{4 \choose 2}{11 \choose 1}{4 \choose 1}}{{52 \choose 5}}.$$ My logic was that I first choose each rank separated and then their respective pairs from the rank. I also know that I am overcounting but I do not know why I am wrong specifically. Any help works!

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Notice that if you use ${13 \choose 1} {12\choose 1}$ to count the number of distinct pairs, you are considering that order matters, that is, King and Queen, for instance, will be counted twice (K, Q) and (Q, K).

As an exercise, prove: ${13 \choose 2} = \dfrac{1}{2} {13 \choose 1} {12\choose 1}.$