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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

3 Answers 3

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Hint: $\displaystyle\binom{n}{n_1,n_2,n_3}=\binom{n}{n_1}\binom{n-n_1}{n_2}$

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Hints:

(1) Calculate the left-hand side of your expression for a few values of $n$, say $1$, $2$, and $3$. Compare with the supposed value of the right-hand side.

(2) Look up the Wikipedia entry for the Multinomial Theorem.

(3) Make a conjecture that is of the same general nature as the problem you were given, but correct, and prove that conjecture. Ideally, the right-hand side should be $1$.

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Actually the answer is exactly as you suggested it to be $(-1+-1+-1)^n = (-1)^n 3^n$, and not 1.

Table[Sum[
  Multinomial[n1, n2, n3] Boole[n1 + n2 + n3 == n] (-1)^(
   n1 - n2 + n3), {n1, 0, n}, {n2, 0, n}, {n3, 0, n}], {n, 0, 10}]

{1, -3, 9, -27, 81, -243, 729, -2187, 6561, -19683, 59049}

You can easily check this with $n=1$. There are 3 possibly combination $n_1 = n_2 = 0 \land n_3 = 1$, or $n_1=n_3=0 \land n_2 =1$ or $n_2=n_3=0 \land n_1=1$.

$$ \binom{1}{1,0,0}(-1) + \binom{1}{0,1,0}(-1) + \binom{1}{0,0,1}(-1) = (-1)+(-1)+(-1) = -3 \not= 1 $$