Can you give a Combinatorial argument that $$\binom{3n}{3}=3\binom{n}{3}+(3)(2)\binom{n}{2}\binom{n}{1}+\binom{n}{1}\binom{n}{1}\binom{n}{1}?$$
Combinatorial Argument
3
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combinatorics
binomial-coefficients
combinatorial-proofs
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2I 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. – 2011-12-07