Problem: In the card game bridge the game begins by dividing up a deck of $52$ cards among four players with each receiving $13$. (We can assume that the players are numbered from $1$ to $4$.) Find the number of ways Player $1$ and Player $2$ together can receive all the heart cards.
My answer: $\left[{13\choose 1}\cdot{39\choose 12}\cdot{27\choose 1} + {13\choose 2}\cdot{39\choose 11}\cdot{28\choose 2} + {13\choose 3}\cdot{39\choose 10}\cdot{29\choose 3} + \cdots + {13\choose 13}\cdot{39\choose 0}\cdot{39\choose 13}\right]\cdot {26\choose 13}\cdot {13\choose 13}$.
Player $1$ could have $1$ heart and $12$ non-hearts while Player $2$ would have to have $1$ non-heart. Or Player $1$ could have $2$ hearts and $11$ non-hearts while Player $2$ would have to have $2$ non-hearts. And so on. Then of the $26$ remaining cards, Player $3$ could have any $13$ while Player $1$ would have to take the remainder.
Since the number of choices for each player in my sequence does not depend on the choices of the other players, we can use the multiplication principle.
Question: Is my reasoning or answer correct? I mean is there a combinatorial identity or something that simplifies my sum which represents the way the hearts can be distributed among the first two players?
Best Answer
Your answer might work, mostly, but there is an easier way. You seem to not have counted the case where the first hand gets all the hearts (or maybe the second.)
The easier way: There are $\binom{39}{13}$ subsets of the deck of size $26$ and containing all the hearts. For each of those subsets, there are $\binom{26}{13}$ ways to distribute those cards to Player 1 and Player 2. And there are $\binom{26}{13}$ ways to divy up the rest of the cards to players 3 and 4.
So we've stumbled on the identity:
$$\sum_{i=0}^{13}\binom{13}{i}\binom{39}{13-i}\binom{26+i}{i}=\binom{39}{13}\binom{26}{13},$$ with the left side your count (corrected to include $i=0,$) and the right side is mine, both excluding the common factor of $\binom{26}{13}.$
If $4n$ is the number of cards in the deck, and there are $k$ cards of type $S,$ then the we get an identity:
$$\sum_{i=0}^{k}\binom{k}{i}\binom{4n-k}{n-i}\binom{3n-k+i}{i+n-k}=\binom{4n-k}{2n-k}\binom{2n}{n}.$$
As noted in comments, this identity can be proved by Vandermonde's identity.
Even more generally, if player $i$ gets $c_i$ cards, and $S$ has $k$ cards, and $c=c_1+c_2+c_3+c_4,$ then there are $\binom{c-k}{c_1+c_2-k}$ ways to pick $c_1+c_2$ cards that have all of $S,$ and $\binom{c_1+c_2}{c_1}$ ways of dividing the cards up for players $1$ and $2.$ Your count gives:
$$\sum_{i=0}^{k}\binom{k}{i}\binom{c-k}{c_1-i}\binom{c+i-k-c_1}{c_2+i-k}=\binom{c}{c_1+c_2-k}\binom{c_1+c_2}{c_1}.$$