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?
 A: My approach would be  
For the first hand  


*

*Choose a rank to be in first hand as the three-card rank: $13\choose1$, $12$ of full ranks remaining

*Choose $3$ cards of that rank: $4\choose3$.   

*Choose a rank to be in first hand as the two-card rank: $12\choose1$, $12$ of $2^+$-card ranks remaining, $11$ of $4$-card ranks remaining.  

*Choose $2$ cards of that rank: $4\choose2$.  


First hand total: ${13\choose1}\cdot{4\choose3}\cdot{12\choose1}\cdot{4\choose2}$
Here comes the tricky part:  


*

*Choose a rank to be in second hand as the three-card rank: $11\choose1$, 10 of $4$-card ranks remaining and $11$ of $2^+$-card ranks remaining.  

*Choose $3$ cards of that rank: $4\choose3$.   


Case 1: The two-card rank that appears in first hand is in the second hand


*Choose that rank as the two-card rank: $1$.

*Choose the remaining $2$ cards of that rank: $1$.  


Case 2: The two-card rank that appears in first hand is not in the second hand


*Choose a rank to be in first hand as the two-card rank:$10\choose1$.

*Choose $2$ cards of that rank:$4\choose2$.  


Second hand total: ${11\choose1}\cdot{4\choose3}\cdot\left[1+{10\choose1}\cdot{4\choose2}\right]$
Grand total: ${13\choose1}\cdot{4\choose3}\cdot{12\choose1}\cdot{4\choose2}\cdot{11\choose1}\cdot{4\choose3}\cdot\left[1+{10\choose1}\cdot{4\choose2}\right]=10 \ 048 \ 896
$
A: *

*choose the two three-card ranks. The number  of ways is $\binom{13}{2}\binom{4}{3}\binom{4}{3}$.

*choose a two-card rank. The number of ways is $\binom{11}{1}\binom{4}{2}$.

*the next two-card rank may be the same or different as the first one. The number of ways is $\binom{2}{2}+\binom{10}{1}\binom{4}{2}$

*Take the two players into account, the number of ways is
$$2\binom{13}{2}\binom{4}{3}\binom{4}{3}\binom{11}{1}\binom{4}{2}\left[\binom{2}{2}+\binom{10}{1}\binom{4}{2}\right]$$
A: You should be multiplying the cases because for each first person full house you have many possibilities for the second person full house.  There are fewer second person full houses available because there are only $11$ ranks for the three of a kind.  You need to condition on whether the pairs match because there are a different number of ways to select the pair cards in the two cases.
A: 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.
