Probability of getting two pair in poker I was looking at this website http://www.cwu.edu/~glasbys/POKER.HTM and I read the explanation for how to calculate the probability of getting a full house.  To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.
In other words, for a full house $P=$
$$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$
Following this logic, I tried to calculate the probability of getting two pair.  My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.  
So I tried $P=$
$$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
Obviously my solution was incorrect.  I read explanation and the correct answer is $P=$
$$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
I'm still a bit fuzzy on where I went wrong though.  Can anyone help me understand this problem a little better?  Thank you very much for your help.

I find permutation more intuitive to follow for this kind of problems. For people like me:
We have five slots to fill: - - - - - . The first slot can take all 52 cards. The second slot can take only three cards so that they can make a pair. Similarly, the third and fourth slots can take 48 and 3 cards, respectively. The last and final slot can take any of remaining 44 cars. Therefore:
52 * 3 * 48 * 3 * 44 = 988416. Please note, this is order dependent. In other words, this is the count of x x y y z. However, we should count all the possibilities (i.e., z x y x y). Therefore, we multiply 988416 with 5! and divide by 2! (order between two xs) * 2! (order between two ys) and 2! (order between the pair of xs and ys). The total count is 14826240.
This is the numerator. The denominator is 52*51*50*49*48 = 311875200. The probability is 0.0475390156062425.
Note that if you want to count how many different hands can be dealt, then you have to divide 14826240 by 5! to compute the combination.
