Need help understanding poker hand probability/combinatorics. I have some trouble understanding a difference when calculating the probability for a full house versus two pairs.
For two pairs the number of possible hands is:
$$ \binom{13}{2} \binom{4}{2} \binom{4}{2} \binom{11}{1}\binom{4}{1} $$
Which makes sense for me as we first choose 2 values out of 13 possible, then 2 suits out of 4 possible twice to get the two pairs, then we draw the last card.
However, when applyng this logic to a full house, I encounter a problem.
For a full house the number of possible hands is:
$$ \binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2} $$
Here we first choose 1 value out of 13 possible and 3 suits of 4 possible to get the three of a kind. Then we choose 1 value out of 12 possible and 2 suits out of 4 possible to get the pair.
Why wouldn't we choose 2 values out of 13 possible, then 3 suits out of 4 possible to get the three of a kind then 2 suits out of 4 possible to get the pair like this?
$$ \binom{13}{2} \binom{4}{3} \binom{4}{2} $$
Or the other way around, apply the full house method to the two pairs:
$$ \binom{13}{1} \binom{4}{2} \binom{12}{1} \binom{4}{2} \binom{11}{1}\binom{4}{1} $$
I hope someone can help me understand the inconsistency in these two calculations.
 A: You can't apply the two pair method to the full house because the 2 "ranks" out of 13 that you choose are not symmetric like a two pair.
Consider a two pair with 5's and 4's
5-5-4-4-X 
If you switch the 5's and the 4's the hand would stay the same
However, for a full house you choose one "rank" to be the 3 of a kind and one rank to be the pair
5-5-5-4-4 is different from 4-4-4-5-5
By your "two pair method" applied to a full house you don't distinguish the above two hands so you undercount.

Now, you could apply the "full house method" to two pairs, but with a few changes.
If you apply this method to two pairs, 5-5-4-4-X would be distinguished from 4-4-5-5-X, which is not desired. (If 5 is chosen from the first selection then 4, and vice-versa)
So every two pair will be counted twice, so you need to divide by two
"Two pair method" for two pairs: $\binom {13}{2}\binom {4}{2}\binom {4}{2}\binom {11}{1}\binom {4}{1} $
This is just a factor of two away from the "full house method" for two pairs:
$\binom {13}{1}\binom {4}{2}\binom {12}{1}\binom {4}{2}\binom{11}{1}\binom {4}{1}$, because:
$\binom{13}{2} = \frac{1}{2} * \binom{13}{1} \binom{12}{1}$
Hope this helps!

Note: (Another way to think about the two pair) For the last card in the two pair, you don't need to do $\binom{11}{1}\binom{4}{1}$ explicitly. As 5's and 4's are ruled out, you can choose any one of the 52-8 = 44 remaining cards (which is the same as $\binom{11}{1}\binom{4}{1}$)
