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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 $f+2g=12$.

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 $f+2g=12$.

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

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

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 $f+2g=12$.

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 $f+2g=12$.

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

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

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 $f+2g=12$.

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

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

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 $f+2g=12$.