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5 deleted 48 characters in body

Warning: This is a pretty ugly answer.

Label the players as 1,2,3, and 4. We first count the total possible number of hands that 1 can be dealt. Let $a$ be the number of singletons 1 has and let $b$ be the number of duplicates 1 has. This yields a total number of

$$\sum \binom{24}{a} \binom{24-a}{b}$$ possibilities, where the sum ranges over all $a$ and $b$ such that $a+2b=12$.

We now count the total number of possibilities for 1 and 2 (by conditioning on what 1 has first). We let $c$ denote the number of singletons that both 1 and 2 have. We let $d$ denote the number of singletons that 2 has but 1 does not have. Finally we let $e$ denote the number of duplicates that 2 has. Then the total number of possibilites for $1$ and $2$ is

$$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e}$$

possibilities, where the sum ranges over all $a, b,c,d, e$ such that $a+2b=12$, $c \leq a$, and $c+d+2e=12$.

We end by computing the total number of possibilities for 1,2, and 3. This actually gives all possibilities since then the hand for player 4 is determined. Let $f$ and $g$ denote the number of singletons and duplicates for 3 respectively. The total answer is then

$$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e} \binom{24-a-b-d-e}{g}\binom{24-b-c-e-g}{f}$$

where the sum ranges over all $a, b,c,d, e, f$ and $g$ such that $a+2b=12$, $c \leq a$, $c+d+2e=12$, and $f+2g=12$.

4 added 2 characters in body

Warning: This is a pretty ugly answer.

Label the players as 1,2,3, and 4. We first count the total possible number of hands that 1 can be dealt. Let $a$ be the number of singletons 1 has and let $b$ be the number of duplicates 1 has. This yields a total number of

$$\sum \binom{24}{a} \binom{24-a}{b}$$ possibilities, where the sum ranges over all $a$ and $b$ such that $a+2b=12$.

We now count the total number of possibilities for 1 and 2 (by conditioning on what 1 has first). We let $c$ denote the number of singletons that both 1 and 2 have. We let $d$ denote the number of singletons that 2 has but 1 does not have. Finally we let $e$ denote the number of duplicates that 2 has. Then the total number of possibilites for $1$ and $2$ is

$$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e}$$

possibilities, where the sum ranges over all $a, b,c,d, e$ such that $a+2b=12$, $c \leq a$, and $c+d+2e=12$.

We end by computing the total number of possibilities for 1,2, and 3. This actually gives all possibilities since then the hand for player 4 is determined. Let $f$ and $g$ denote the number of singletons and duplicates for 3 respectively. The total answer is then

$$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e} \binom{24-a-b-d-e}{g}\binom{24-b-e-g}{f}$$binom{24-a-b-d-e}{g}\binom{24-b-c-e-g}{f}$$ where the sum ranges over all a, b,c,d, e, f and g such that a+2b=12, c \leq a, c+d+2e=12, and f+2g=12. 3 deleted 2 characters in body Warning: This is a pretty ugly answer. Label the players as 1,2,3, and 4. We first count the total possible number of hands that 1 can be dealt. Let a be the number of singletons 1 has and let b be the number of duplicates 1 has. This yields a total number of $$\sum \binom{24}{a} \binom{24-a}{b}$$ possibilities, where the sum ranges over all a and b such that a+2b=12. We now count the total number of possibilities for 1 and 2 (by conditioning on what 1 has first). We let c denote the number of singletons that both 1 and 2 have. We let d denote the number of singletons that 2 has but 1 does not have. Finally we let e denote the number of duplicates that 2 has. Then the total number of possibilites for 1 and 2 is $$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e}$$ possibilities, where the sum ranges over all a, b,c,d, e such that a+2b=12, c \leq a, and c+d+2e=12. We end by computing the total number of possibilities for 1,2, and 3. This actually gives all possibilities since then the hand for player 4 is determined. Let f and g denote the number of singletons and duplicates for 3 respectively. The total answer is then $$\sum \binom{24}{a} \binom{24-a}{b} \binom{a}{c} \binom{24-a-b}{d} \binom{24-a-b-d}{e} \binom{24-a-b-d-e}{g}\binom{24-b-d-e-g}{f}$$binom{24-a-b-d-e}{g}\binom{24-b-e-g}{f}$$

where the sum ranges over all $a, b,c,d, e, f$ and $g$ such that $a+2b=12$, $c \leq a$, $c+d+2e=12$, and $f+2g=12$.

2 deleted 128 characters in body; deleted 4 characters in body
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