# From $3$ red, $4$ green and $5$ yellow balls, how many selections consisting of $6$ balls are possible, if each color must be represented twice?

From $3$ red, $4$ green and $5$ yellow balls, how many selections consisting of $6$ balls are possible, if each color must be represented twice?

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Could you provide some background why you are interested in this question. If it is a homework, please include the homework task. Also indicate if you did already try something? Maybe with a smaller number of balls? –  Fabian Nov 13 '12 at 14:33
It is an exercise in an old textbook (Statistics: A Foundation for Analysis by Hughes and Grawoig). According to the answers there should be 180 selections consisting of 6 balls with each color having two representations. Because there are several like objects, the total number of permutations should be 27,720. –  Santiago Bueno Nov 13 '12 at 16:10

That question just makes sense if the balls of same color are distinguishable. If not, you have just 1 possible selection, taking 2 balls of each color.

So if they are distinguishable, you have $\binom{3}{2}$ options of choosing the red balls, $\binom{4}{2}$ of choosing the green, and $\binom{5}{2}$ of choosing the yellow. That are in total $3\cdot6\cdot10=180$ possible ways of choosing the balls.

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Yeeeesssss, your way of computing it is the right way. There is a little bit of a mistake since 5C2 = 10, not 9. The answer is supposed to be 180. –  Santiago Bueno Nov 13 '12 at 16:20

Select two balls from each color; multiply the numbers of possibilities for each color together.

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You're assuming that the balls of the same color are distinguishable? I guess this works even if you don't: $1\cdot1\cdot1=1$ –  robjohn Nov 13 '12 at 15:40