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Help me please with the following question:

Let $n \in N$ be fixed. I would like to find an upper and lower bound for the following function:

$$ f(x)={2n-x \choose n}+{x \choose 2}{2n-x \choose n-2}+\ldots+{x \choose n}-{x \choose 1}{2n-x \choose n-1}-{x \choose 3}{2n-x \choose n-3}-\ldots -{x \choose x-1}{2n-x \choose 1}. $$

I know that I have to find for which $x$, $f'(x)=0$, but I don't know how to differentiate this function.

Thank you.

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Are you looking for a maximum or just an upper bound? – wj32 Nov 1 '12 at 22:19
The way you've written $f(x)$ only makes sense for even $n$; otherwise it's not clear how the dots are to be interpreted. – joriki Nov 2 '12 at 0:16
You're choosing $n$ items from $2n$ items, distinguished according to how many you choose from some particular subset $X$ of $x$ items, and you're considering the excess of selections with an even number of items in $X$ over selections with an odd number of items in $X$. For $x=0$ or $x=2n$, all selections have an even number of items in $X$, so the best upper bound is $\binom{2n}n$. A lower bound is more difficult to obtain; I would suspect that it's attained at $x=n$, and $\sum\binom nk\binom n{n-k}(-1)^k=\sum\binom nk^2(-1)^k$ can be positive or negative. – joriki Nov 2 '12 at 1:18

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