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An urn contains 4 red, 5 blue and 3 green marbles. Six are drawn at random (without replacement) one at a time. How many ways could the sequence of six marbles contain two of each color?

I thought that it would be


but that gives me 180 and the answer is 90.

Any help would be highly appreciated!

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Your answer counts the number of different sets of six marbles (two of each color) that could be drawn, assuming the marbles can be distinguished from each other: Choose two of the red ones, two of the blue, and two of the green. In the question you have to answer, though, marbles of the same color are probably assumed to be indistinguishable. In addition, the order of the draw matters. If so, the quantities 4, 5, and 3 are a red herring - the question is really this: How many different six-letter words can be made from two Rs, two Bs, and two Gs? – Steve Kass Mar 26 '14 at 22:20
Thank you very much for your response. I'm glad I got 2 different perspectives on this problem. – WhatsAGuitar Mar 26 '14 at 22:36
up vote 1 down vote accepted

First, there are $\binom{6}{2}$ ways how to select the red marbles out of the sequence of 6 drawn marbles. Then you have 4 marbles left. You pick 2 of them to be blue, which gives you then $\binom{4}{2}$ possibilities. The remaining marbles are green - so noting to choose. Thus in total you have $\binom{4}{2}\binom{6}{2}=90$ sequences.

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Thank you so much! I never thought of it that way. Thank you very much for your help. – WhatsAGuitar Mar 26 '14 at 22:33

Six distinct marbles can be sorted $6!$ ways, but when we have three pairs the ordering within each pair doesn't matter, so the multiset's permutation is:

$$\frac{6!}{2!2!2!} = 90$$

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