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Given 4 blue cards, 3 red cards and 2 green cards, in how many possible ways one can hand them out to two different players?

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    What are your thoughts? How many ways are there to distribute the green cards, say? – 2017-01-06
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    ${3 \choose 2}$ for the green, ${4 \choose 2}$ for the red, ${5 \choose 2}$ for the blue? – 2017-01-06
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    Not following you. For the green, I can give player $A$ either $0,1,2$ so $3$. True, that happens to equal $\binom 32$ but I think that's an accident. For the red, I can give $A$ any of $0,1,2,3$ so $4$ cases (whereas $\binom 42=6$) – 2017-01-06
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    my bad, I get your point. so there are 3 for the green, 4 for the red and 5 for the blue? and does multiplying them give me the answer? – 2017-01-06
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    That is correct. – 2017-01-06

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Note that $3$ ways to split the two green cards between the two players $(P_1,P_2)$: $${(0,2),(1,1),(2,1)}$$ Likewise, there will be $4$ ways to split the three green cards:

$${(0,3),(1,2),(2,1),(3,0)},$$

and $5$ ways to split the four blue cards:

$${(0,4),(1,3),(2,2),(3,1),(4,0)}.$$

Thus $3\cdot 4\cdot 5=60$ is the number you are looking for.

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    thanks. what if I wanted to deal them to n different people? – 2017-01-06
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    Count the number non-negative integer solutions to the equation $$x_1+x_2+\cdots+x_n=k$$ where $k$ is the number of cards of a given color. Do the same for each color, and then multiply the resuts. – 2017-01-06
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    ps. The number of solutions the to equation above is $\binom{k+n-1}{k}$. – 2017-01-06