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The problem asks us to calculate:
$ \sum_{i = 0}^{n}(-1)^i \binom{n}{i} \binom{n}{n-i}$

The way I tried solving is:
The given sum is the coefficient of $x^n$ in $ (1+x)^n(1-x)^n $, which is $ (1 - x^2)^n $. The coefficient of $x^n$ in $(1 -x^2)^n$ is $ (-1)^{n/2} \binom{n}{n/2} $.

Am I doing it right?

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Thanks, fixed it. –  user813 Sep 22 '10 at 15:57

1 Answer 1

up vote 4 down vote accepted

Your solution is correct for even $n$.

If $n$ is odd then your last sentence should read "The coefficient of $x^n$ in $(1-x^2)^n$ is $0$." This is because only even powers of $x$ occur when expanding $(1-x^2)^n$.

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Ah, good. Thanks. –  user813 Sep 22 '10 at 16:01

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