<s>[INST] Counting card distributions when cards are duplicated If we have a deck of 48 different cards and 4 players each get 12 cards, it is well known how to calculate the number of possible distributions: take fac(48) and divide 4 times by fac(12).
In a german card came (Doppelkopf) there are 24 different card types, but 2 copies of each type, that is 48 cards at all.
How many distributions when there are 4 players? My first approach was to assume there are 48 different cards, and then to divide by 2 and by 2 (... 24 times). But this underestimates the real number of distributions. Because: Say player 1 has both spade kings. This has been ruled out already by dividing by fak(12). But if player 1 and player 2 both have a spade king, then we must divide by 2.
How to calculate the number of different distibutions?
Is there a closed formula as in the single card deck case?
 [/INST] 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 $f+2g=12$.
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