When drawing a pair the number of ways to get it is:
$${13 \choose 1}{4 \choose 2}{12 \choose 3}(4)^3$$
But I don't understand why the following doesn't work: $${13 \choose 1}{4 \choose 2}{12 \choose 1}{11 \choose 1}{10 \choose 1}(4)^3$$
Thanks a lot!
Btw This hand has 5 cards.
On
Basically, your question here is what's the difference between $\binom{12}{3}$ and $\binom{12}{1} \binom{11}{1} \binom {10}{1}$. The difference is that the latter implicitly orders what you're picking. In this instance, you're picking the 3 cards in your hand that aren't part of the pair.
The former way of counting just picks three card values (e.g. 5, Q, A, not part of your pair, say K). The latter counts how many ways can I pick the first card, the second, and the third. Thus, while the first counts 5, Queen, Ace once, the latter counts it 6 times, once for each permutation of those values.
Notice that the second way differs from the first by a factor of $3!$. This represents the fact that the other three cards can be drawn in any order. The second way is counting as two possibilities drawing the other three cards in two different orders, which is not what you want.
Edit: OP asked why we're not concerned with double counting when we simply multiply the "pair" part by the "other cards" part. Notice that in this case we're not double counting anything: it's not as if we've counted twice the possibility of JJ543 by multiplying. One might say that the pair is distinguishable from the other three cards, so there's no question of double counting them in different orders.
The trick with these things is to convince yourself in your mind that your formula has created a bijection between the possibilities and the things that you're counting. Instead of memorizing rules about order, etc., I would rather practice coming up with the bijection in your head, and checking that you haven't missed anything. Think "for each hand with a pair, there's a unique term in my counting, and for each term in my counting, there's a unique hand with a pair."