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Can you give a Combinatorial argument that $$\binom{3n}{3}=3\binom{n}{3}+(3)(2)\binom{n}{2}\binom{n}{1}+\binom{n}{1}\binom{n}{1}\binom{n}{1}?$$

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I can. You should also be able to. We want to choose $3$ people from $3n$ people, $n$ with brown eyes, $n$ with blue eyes, $n$ with red eyes. We can choose all with brown, $\binom{n}{3}$ ways, or same blue, same red, total $3\binom{n}{3}$, your first term. We can choose $1$ of each eye colour, that's your last term. You can figure out what your middle term counts. –  André Nicolas Dec 7 '11 at 15:39

1 Answer 1

You have three urns, each with $n$ objects in it. The number of ways to select three objects total from the urns is the left hand side ($ ({3n\atop3})$) .

The right hand side consists of three terms: the first term ($3({n\atop3})$) is the number of ways to select the three objects with all three in the same urn. The second term ($3\cdot2\cdot({n\atop2})({n\atop1})$) is the number of ways to select the three objects by first picking an urn and selecting two objects, and then selecting one of the two remaining urns and selecting an object. The last term ($({n\atop1})({n\atop1})({n\atop1})$) is the number of ways to select the three objects with one selected from each urn. Adding these together gives you the total number of ways to select three objects from the urns.

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Thank you both, this is very helpful. I was stuck on this problem and I guess I was just thinking about it too much. I was thinking it was a stars and bars problem. –  Mark Dec 7 '11 at 15:58
    
Salamat po @J.M. for correcting my question. –  Mark Dec 7 '11 at 16:13

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