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In how many ways can a collection of 10 marbles be formed from Red, Yellow and Blue marbles? When only the number of Red, Yellow and Blue marbles matter, not the order in which they are drawn.

I am thinking this has something to do with a ball and box problem? I thought that taking 3^10 and dividing it by 10! would work but the answer is a pretty big fraction.

Could someone please help me? Which are the balls and which is the boxes?

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Not sure exactly how you want to apply balls and boxes here, but it looks like you can turn this into a stars and bars problem with 10 stars and 2 bars.

The idea here being that the "red marbles" would be the stars to the left of both bars, the "yellow marbles" would be stars between the two bars, and the "blue marbles" would be stars to the right of both bars.

I'll let you work out the exact math from there.

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  • $\begingroup$ So I was thinking using the formula n+k+1 choose k-1. Do you think I'm heading in the right direction? $\endgroup$ – A Glenn Nov 15 '13 at 23:40
  • $\begingroup$ Close. (The article already has the formula, just fyi) $\endgroup$ – Dennis Meng Nov 16 '13 at 0:06
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Use stars and bars. All configurations can be mapped to a sequence of stars and bars as in the following example.

$$3R,4Y,3B\to RRRYYYYBBB\to RRR\vert YYYY\vert BBB\to ***\vert****\vert***$$

How many such sequences of stars and bars are there?

A remark on your method: you cannot eliminate the overcounting due to different orderings by dividing by $10!.$ The reason is that same colored balls are not distinguishable from each other but differently colored ones are. For my example above, there are $\dfrac{10!}{3!\,4!\,3!}$ orderings, but with a different color composition, the number of orderings would be different. Hence there is no single overcounting factor that you can divide by.

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