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.
 A: The problem with your $247,104$ is that it counts each two-pair hand two times, according to which of the pairs you mention first. But 5H-5D-7S-7H-9D is the same hand as 7S-7H-5H-5D-9D, so it gets counted both with fives first and with sevens first.
In contrast, for a full house it is unambiguous which value is the one with three cards and which is the one with two.
A: 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:$\binom{11}{1}\binom{4}{1}$
A: In the second formula you give for the number of was to draw a two pair which is a product of $6$ terms, the outcomes (4, heart and spade, 3, heart and diamond, J, club) and (3, heart and diamond, 4, heart and spade, J, club) are counted as different hands when in fact they are the same five-card hand.  So you need to divide this value by $2$ to get the correct answer.  
In the second formula you give for the number of ways to draw a full house (your formula should evaluate to $1872$, not $2808$),  the first term ${13 \choose 2}$ is the number of ways to draw two kinds from $13$ kinds, and from these two kinds, you can choose the kind for the triple in $2$ ways (recall that a full house has both a triple and a pair).  So you need to multiply your answer $1872$ by $2$.  Here, since the order in which you draw the two kinds is important, using combinations (binomial coefficient) just once will not suffice.  
The reason you are using a different method for the two problems is that in a full house, the two subsets of cards have different sizes ($2$ and $3$) whereas if the hand is two pairs, then the two subsets of cards have the same size $2$. You are correct in thinking of the experiment as being a sequence of steps and taking the product of the number of ways to do each step, but during this process you need to be careful to not over-count or under-count.
