# Clarification for this combinations/permutations problem?

I've been going through a list of poker hands and their descriptions, and then attempting to calculate their probabilities by first calculating the number of possible hands for the given hand.

I tried to do the Two Pair hand, which is a hand where you have 2 cards of the same value, and another 2 cards of the same value but different from the previous pair, and one card of a value different from the pairs(e.g. $3\heartsuit3\spadesuit \space 4\clubsuit$ $4\spadesuit \space10\heartsuit$); I got the wrong answer:

But why is my approach wrong? I thought of it as choosing a value out of 13 possible values(A,K,J,10,...), then choose 2 cards; choose another value from the remaining 12 values, then another 2 cards. But this isn't the same as choosing a pair out of 13 values, and then choose 4 cards as it appears in the correct answer. I can't see how that makes a difference intuitively... what's the difference here?

Your first way double counts. It counts two Jacks, and two $5$'s, and some useless card, as different from two $5$'s, and two Jacks, and some useless card.

For Jack is one of the $\binom{13}{1}$ kinds that you chose "first," and $5$ is among the $\binom{12}{1}$ kinds that you chose "second." But $5$ is among the $\binom{13}{1}$ kinds that you chose "first," and Jack is among the $\binom{12}{1}$ kinds that you chose "second."

Note that this issue does not arise with a full house, for three Jacks and two $5$'s is a different hand than three $5$'s and two Jacks.