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Prove that: $\sum_{n_1+n_2+n_3 = n} \binom{n}{n_1,n_2,n_3} \cdot (-1)^{n_2} = 1$

The sum is over all positive integer solutions of $n_1 + n_2 + n_3 = n$.

Use the multinomial theorem:

$(x_1 + x_2 + x_3)^n = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} x_1^{n_1} x_2^{n_2} x_3^{n_3}$

Let $x_2 = z$, $x_1 = x_3 = 0$ and plug into the above:

$(0 + z + 0)^n = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} 0^{n_1} z^{n_2} 0^{n_3} = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} z^{n_2}$

Let $z = -1$ and plug in:

$\sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} (-1)^{n_2}$

But can't $n_2$ be both even and odd, so it could be either $1$ or $-1$?

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    Note that the mistake you make is saying $0^{n_1} = 0^{n_3} = 1$ which is quite false.2011-09-28

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HINT: From the multinomial theorem, we have

$(x_1 + x_2 + x_3)^n = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} x_1^{n_1} x_2^{n_2} x_3^{n_3}$

Now try plugging in $x_1 = x_3 = 1$ and $x_2 = -1$...

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    You can let $x_1 = x_3 = 0$ any time you want. It just doesn't give you what you want, neither on the left nor on the right. You want $x_1^{n_1} = 1$ and $x_3^{n_3} = 1$ and $x_2^{n_2} = (-1)^{n_2}$, so only one choice of $x_1,x_2,x_3$ is suitable.2011-09-28