How many possibilities are there for two full houses to be dealt to two players in one game? Imagine dealing cards from a classic 52 card deck to two poker players.
How many possibilities are there for both of them to be dealt a full house(three cards in same rank and two cards of another rank) in same round? 
As I know totally ${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}$ possibilities for one player (AAABB). According to that, I tried to solve my problem as shown below: $${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 1}{4 \choose 2} + {13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 3}{11 \choose 1}{4 \choose 2}$$ for two players ((AAABB)(AAACC) or (AAABB)(CCCBB)). Is my solution correct or am I missing something?

There are two possibilities.  Either both players receive two cards of the same rank or they do not.
Both players receive two cards of the same rank:  There are $\binom{13}{1}$ ways to choose the rank from which player A receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank.  That leaves $\binom{12}{1}$ possible ways to choose the rank from which player B receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank.  There are $\binom{11}{1}$ ways to choose the rank from which both players receive two cards.  There are $\binom{4}{2}$ ways to select which two of the four cards are received by player A.  Hence, there are 
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}$$
such deals.
The players receive two cards from different ranks:  The number of ways to select three cards of one rank for each player is the same as above.  There are $\binom{11}{1}$ ways to select the rank from which player A receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank.  There are $\binom{10}{1}$ ways to select the rank from which player B receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank.  Hence, there are 
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}\binom{10}{1}\binom{4}{2}$$
such deals.
Since the two cases are mutually exclusive and exhaustive, the number of deals in which both players receive a full house is found by adding the results for the above cases.
