two pairs in poker

Question

I have tried to solve number of hands that give two pairs in 5 hand poker. From this site:

Probability of getting two pair in poker

the answer is:

$${13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}$$

But I don't get why it is not:

$${13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}$$

That is why it is not 13 ways of drawing pair 1 and then 12 ways to draw pair two instead of $13C2$. Can someone explain this since I often have similar problems in probability. I understand that $13C2$ is the number of ways of selecting two in a pile and not selecting 11 in a pile and that thoose two are the two values of the two pairs. But why does this give the right amount of hands for two pairs?

Answer 1

$13 \choose 2$ selects the two pairs without considering their order. ${13 \choose 1}{12 \choose 1}$ selects two pairs but considers aces and twos different from twos and aces. That is the source of the factor two. If we just want the number of hands, the first is correct as we don't care what order we drew the cards in.