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I really am not sure how to make headway on this exercise.

And what do I have wrong with $(-1)^{n_1-n_2+n_3} = (-1)^{n_1}(-1)^{n_2}(-1)^{n_3} = (-1)^n$? Because from there I would have the statement as equivalent to $(-1 + -1 + -1)^n$, which clearly isn't right.

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    Prove that what? Your sentence is incomplete. Or maybe the title got truncated. That's a good reason to put the full question in the body of the question. Your calculation of $(-1)^{n_1-n_2+n_3}$ is right.2011-10-15
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    I apologize for forgetting an important part of the problem statement. It is now included.2011-10-15
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    If you look at $n=1$, you can see that this can't be right.2011-10-15
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    I'm guessing the problem was supposed to be $\sum_{n_1+n_2+n_3=n}\binom{n}{n_1,n_2,n_3}(-1)^{n_2}$.2011-10-15
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    @Jim: I double checked this time. The problem statement is correct.2011-10-15
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    For $n=1$ we have $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. The multinomial coefficient $\binom{1}{1,0,0}=1$ and similarly for the other two. Also $(-1)^{n_1-n_2+n_3}=-1$ in all three cases, so the total sum is $-3$.2011-10-15
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    The exponent is not right. If it is changed to $(-1)^{n_1}$, that would be OK. There are other versions that would work.2011-10-15

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