Concerning The Number of Ways of Drawing a Full House vs. Two Pair The Wikipedia entry for "Poker probability" gives the following result for the number of ways of drawing a full house:
$$ \binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2} = 3, 744. $$
The logic here is conventional: From the 13 kinds you choose 1, then choose 3 of the 4 available cards of that kind. Then, from the remaining 12 kinds, you choose 1, then choose 2 of the available cards of that kind. No problem.
However, for the number of ways of drawing a two pair it gives the following:
$$ \binom{13}{2} \binom{4}{2} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 123, 552. $$
But, by the logic used for drawing a full house, shouldn't it be the following:
$$ \binom{13}{1} \binom{4}{2} \binom{12}{1} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 247, 104? $$
OR, conversely, by using the logic used to get a two pair, shouldn't the number of ways of drawing a full house be the following:
$$ \binom{13}{2} \binom{4}{3} \binom{4}{2} = 2,808? $$
As you can see, this isn't just a matter of inconsistent approaches that give the same result; these give different values. I just don't know how I'm supposed to be able to intuit these sort of things, seeing as they seem like the same thing to me.

The difference between a full house (AAABB) and a two pair (AABBC) is that when drawing a two pair, both pairs are groups of two. With the full house, one is a group of three, and the other is a group of two. With your methos, you are counting, for example, AABBC as different from BBAAC.
So the logic is first choose the two (different) kinds from a total of $13$:$\binom{13}{2}$；Then for each of these values, pick two out of four:$\binom{4}{2}\binom{4}{2}$;
And from the remaining $11$ kinds, choose the value of the other card:$\binom{11}{1}\binom{4}{1}$
