The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$ $\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}}
\newcommand{\r}[1]{\frac1{x^{#1}}}$
The value of $$\sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{k+10}$$
It is also the coefficient of $x^{10}$ in:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{20}\r{20}}\b{\c{10}x^{10}+\c{11}x^{11}+...\c{30}x^{30}}$$
Adding some terms maybe won't harm:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{30}\r{30}}\b{\c{0}x^{0}+\c{1}x^{1}+...\c{30}x^{30}}$$
Contracting:
$$\b{1-\r1}^{30}\b{1+x}^{30}=\b{x-\r1}^{30}=\sum_{n=0}^{30}(-1)^{n}\c nx^{2n-30}$$
The $20$th term interests me, giving:
$$\c{20}=\c{10}$$
Well are there any other interesting ways?
 A: Vandermonde's Identity yields
$$
\begin{align}
\sum_{k=0}^{20}\binom{30}{k}\binom{30}{k+10}
&=\sum_{k=0}^{20}\binom{30}{k}\binom{30}{20-k}\\[6pt]
&=\binom{60}{20}
\end{align}
$$

In the product,
$$
\begin{align}
\left(x-\frac1x\right)^{30}
&=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{30-k}x^{-k}\\
&=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{2(15-k)}\\
\end{align}
$$
and in the product
$$
\begin{align}
\left(1-\frac1x\right)^{30}\left(1+x\right)^{30}
&=\sum_{j=0}^{30}(-1)^j\binom{30}{j}x^{-j}\sum_{k=0}^{30}\binom{30}{k}x^k\\
&=\sum_{n=-30}^{30}\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}x^n
\end{align}
$$
Comparing the coefficients of $x^n$, we have for $n$ odd
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}=0
$$
and for $n$ even
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}=(-1)^{15-\frac n2}\binom{30}{15-\frac n2}
$$

For $n=10$, we get
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+10}=\binom{30}{10}
$$
For $n=-10$, we get
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j-10}=\binom{30}{20}
$$
A: Suppose we are interested in the value of
$$S(n,m) = 
\sum_{k=0}^n (-1)^k {n+m\choose k} {n+m\choose k+m}.$$
Introduce
$${n+m\choose k+m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{k+m+1}} \; dz.$$
This  integral controls  the range  being zero  when $k>n$  so  we can
extend the summation to infinity to obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{m+1}}
\sum_{k\ge 0} {n+m\choose k} 
(-1)^k \frac{1}{z^k} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{m+1}}
\left(1-\frac{1}{z}\right)^{n+m} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(z^2-1)^{n+m}}{z^{n+2m+1}} \; dz.$$
Now if $n+2m$  is odd i.e. $n$ is odd we  are extracting a coefficient
from a polynomial that  is even in $z$, so the sum  is zero. If $n$ is
even we get
$$[z^{n+2m}] (z^2-1)^{n+m}
= (-1)^{n/2} {n+m\choose n/2 + m}.$$
With $n=20$ and $m=10$ this yields
$${30\choose 20}.$$
