Difference between full house and two pair in consideration of numbers chosen? So I am a bit confused about why the number of ways to get a full house is:
$ 13 \choose 1 $$ 12 \choose 1$$ 4 \choose 3$$ 4 \choose 2$
as opposed to: 
$ 13 \choose 2 $$ 4 \choose 3$$ 4 \choose 2$. 
Alternatively, with 2 pairs, why it is: 
$ 13 \choose 2 $$ 4 \choose 2$$ 4 \choose 2$$4 \choose 1$$11 \choose 1$
over: 
$ 13 \choose 1 $$ 12 \choose 1 $$ 4 \choose 2$$ 4 \choose 2$$4 \choose 1$$11 \choose 1$
 A: In two pair hands, we need to avoid double counting hands by switching the places of the first two pairs. Say, $(\textrm{Ah Ad Kh Kd Qh})$ and $(\textrm{Kh Kd Ah Ad Qh})$ are the same hand. With full houses, there is no such symmetry, i.e., $\textrm{AAAKK}$ and $\textrm{KKKAA}$ are not the same hand (with any suits of the cards).
The equation for the number of full houses you give can be derived as follows:

*

*The rank of the trips can be selected in $13 \choose 1$ ways.

*Given (1), the rank of the pair can be selected in $12 \choose 1$ ways.

*Given (1,2), the suits of the trips can be selected in $4 \choose 3$ ways.

*Given (1,2,3), the suits of the trips can be selected in $4 \choose 2$ ways.

Your alternative equation with the factor $13 \choose 2$ would in place of 1,2 just count the number of ways to select the rank of the trips and the rank of the pair without caring about which is which. This would "half-count" the number of hands (by considering e.g. selection of $\textrm{AAAKK}$ and $\textrm{KKKAA}$ to produce the same hand in the first step).
For two pair hands, the number of hands can be obtained as follows:

*

*The ranks of the pairs can be selected as $13 \choose 2$ ways

*Given 1, the rank of the fifth card can be selected in $11 \choose 1$ ways

*Given (1,2), the suits of the higher pair can be selected in  $4 \choose 2$ ways

*Given (1,2,3), the suits of the lower pair can be selected in $4 \choose 2$ ways

*Given (1,2,3,4), the rank of the fifth card can be selected in $4 \choose 1$ ways

If we replace the first factor by your alternative, ${13 \choose 1}\, {12 \choose 1}$, this would count the number of ways of selecting the rank of the first pair and then the number of ways of selecting the rank of the second pair. But, there is no first and second pair, so this would double-count the number of hands (by considering, e.g., selection of $\textrm{AAKK}$ and $\textrm{KKAA}$ in the first step to be different).
An exercise
Similar thing occurs for example in 3-card hands when counting the number of pairs (pair and third card are distinct) and the number of no-pairs (symmetry between the three different ranks). To further convince yourself, you might derive similar equations for the number of pair and no-pair 3-card hands with some small deck (say, 3 ranks and 2 suits per rank) where you can also enumerate all the combinations by hand.
