So I'm looking at Wolfram's Probability of Poker Hands page and I don't completely understand the one for two pairs. I understand and agree with everything except division by 2!.

I figure: $\displaystyle (13)$$\displaystyle \binom{4}{2}$$\displaystyle (12)$$\displaystyle \binom{4}{2}$ $\displaystyle 44$ tells us the number of ways to get two of a kind for thirteen face-card values, then two of a kind for the remaining twelve face-card values, and then 44 cards which are none of these two face cards. But this would represent getting one pair first, then the other pair next, then the onesy card last. To correct for this sort of double-counting, we divide. But I don't get why we divide by 2!. I mean, there's the first pair, the second pair, and onesy. Three positions, so to speak. So shouldn't we divide by 3!?

(Note I: I asked about calculating the probability of two pair before, but my question was different--more about the general issue of ordering rather than the particular issue in this example.)

(Note II: What is wrong with this LaTeX math code? It keeps giving me a compile error. 13 \binom{4}{2} \cdot 12 \binom{4}{2} \cdot 44)