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In how many ways can we distribute 20 cards between 2 people given there are 3 hearts for each 2 spades?

Any help would be appreciated.

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Why are using (EXAM TOMORROW!) in headline? We don't need to know that to answer your question and neither does it changes your chance to get an answer from us. – Quixotic Feb 28 '12 at 7:09
Well, good luck! – robjohn Feb 28 '12 at 7:19

Let the four people be N,E,S, and W. Do the problem first for one rank, say the aces. They come in four distinguishable suits, $\spadesuit,\heartsuit,\diamondsuit$, and $\clubsuit$. If you list the four aces in the order in which they appear in the hands of N,E,S, and W, there are $4!=24$ possible orders. In other words, you can distribute the aces in $4!$ different ways.

The reasoning is exactly the same for any other rank, so each rank can be distributed in $4!$ ways. The choices are made independently for each rank, so they must be combined by multiplication $-$ what’s sometimes called the Chinese menu principle. That is, there are $4!=24$ ways to deal out the aces, $4!=24$ ways to deal out the $2$’s, ..., and $4!=24$ ways to deal out the kings. That’s $13$ ranks, so there are $(4!)^{13}=24^{13}=2^{39}3^{13}$ ways to deal out the whole deck.

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Do you play bridge, by any chance? – Aryabhata Feb 28 '12 at 7:30
@Aryabhata: I played a little in grad school, and once in a blue moon I read the bridge column in the local paper, but I’ve not played in over 35 years. – Brian M. Scott Feb 28 '12 at 7:48
I see. I guessed because of the N,E,S,W, which I supposed you already figured out :-) – Aryabhata Feb 28 '12 at 8:40
@Aryabhata: You might also have guessed because of the order in which I listed the suits. :-) – Brian M. Scott Feb 28 '12 at 8:48
Very true ...:-) – Aryabhata Feb 28 '12 at 18:38

Note: Below is a solution of the problem first posted. Thirteen cards were deal from a standard deck to each of $4$ people. In how many ways can each person end up with a hand that has a card of each kind?

Give the people interesting names, such as A, B, C, and D.

How many possible hands are there for A? She could get any of the Aces, and any of the Kings, and any of the Queens, and so on. The number of choices for the Ace is $4$, For each such choice, there are $4$ choices for the King, for a total of $(4)(4)$ so far. For each possible choice of Ace and King, there are $4$ choices for the Queen, for a total so far of $(4)(4)(4)$. Continue. There are $4^{13}$ possible hands for A.

For every hand that A got, B could get any of the $3$ remaining Aces, and any of the $3$ remaining Kings, and so on, for a total of $3^{13}$ possible hands. Thus the number of choices for the hands that A and B get is $4^{13}3^{13}$. For every one of these, there are $2^{13}$ ways to choose the hand that $C$ gets, for a grand total of $$4^{13}3^{13}2^{13}.$$

Another way of thinking about it is that there are $4!$ ways to distribute the Aces between our four people. For every one of these ways, there are $4!$ ways to distribute the Kings, and for every one of these ways there are $4!$ ways to distribute the Queens, and so on for a total of $$(4!)^{13}.$$

Added: The OP's edit has completely changed the problem. There are $3$ $\heartsuit$ for every $2$ $\spadesuit$, so for every $5$ cards, there are $3$ $\heartsuit$ and $2$ $\spadesuit$, for a total of $12$ $\spadesuit$ and $8$ $\heartsuit$. We assume that there are no duplicates among the $\spadesuit$, nor among the $\heartsuit$.

All this is irrelevant if we distribute the cards $10$ to each person. Then there are $\binom{20}{10}$ ways to choose A's hand, so $\binom{20}{10}$ ways to distribute the cards.

If A could get any number of cards, and B the rest, the question is how many subsets our set of cards has. The answer is $2^{20}$. Again, the distribution into $\spadesuit$ and $\heartsuit$ is irrelevant.

Or else we want to give $10$ cards to each person, but we are only interested in the number of $\spadesuit$ and $\heartsuit$ that each person has, but not the actual cards. Then A could have any number of $\spadesuit$ from $0$ to $10$, so there are $11$ ways.

Or else maybe A and B need not get the same number of cards, and we are only interested in how many of each suit each gets. One could go on.

The point is that the new problem is incompletely specified. Thus, depending on how we interpret the problem, we can produce wildly different answers.

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