<s>[INST] 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?
 [/INST] 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
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j-10}=\binom{30}{20}
$$
</s>