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I am asked to prove the following sum:

$$\sum_{i = 0}^n (-1)^i\binom{n}{i}\binom{n}{n-i}.$$

Here is what I did to get started.

First, I noticed that it looked somewhat like Vandermonde's Identity, so I kept it in my mind but ultimately I wasn't able to proceed. Instead I rewrote the binomial coefficients

$$\sum_{i = 0}^n (-1)^i\frac{n!}{i!(n-i)!}\cdot\frac{n!}{(n-i)!(n+i)!}$$

Now part of me thinks there might be some relationship between $(n-i)!(n+i)!$ because it looks nice, but ultimately I don't know if there is, nor if this is even the right approach to go down.

OK, thankfully with the help of @Jochen, turns out I wrote it wrong. We will end up with

$$\sum_{i = 0}^n (-1)^i\binom{n}{i}\binom{n}{i}.$$

Now looking at this, I can't remember any identiy that would help with this. Maybe the Binomial Theorem? I am not entirely sure.

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    Hello :) You didn't rewrite correctly, we have $\binom {n}{n-i}=\frac{n!}{(n-i)!(n-(n-i))!}=\frac{n!}{(n-i)!i!}=\binom ni$ – Jochen Sep 27 '21 at 09:55
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    Maybe it helps to compute the coefficents of $(1+x)^n(1-x)^n$ – Jochen Sep 27 '21 at 10:00

1 Answers1

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It's not helpful to rewrite $\binom{n}{n-k}$ as $\binom nk$.

Observe $(x+1)^n(x-1)^n=(x^2-1)^n$. Hence, $$\sum_{k=0}^n\binom{n}{k}x^{k}\cdot \sum_{k=0}^n\binom{n}{n-k}(-1)^{k}x^{n-k}=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}x^{2k}.$$ Compare the coefficients of $x^n$ of the left handside and right handside. One can see, that it is 0 for odd $n$ and $(-1)^{n/2}\binom{n}{n/2}$ for even $n$.

For $2\nmid n$ you can find an easier argument by replacing $k$ with $n-k$: $$2\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{n}{n-k}=\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{n}{n-k}+\sum_{k=0}^n(-1)^{n-k}\binom{n}{n-k}\binom{n}{k}=\sum_{k=0}^n((-1)^k+(-1)^{n-k})\binom{n}{k}\binom{n}{n-k}=0$$

Jochen
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