The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$

Question

$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}} \newcommand{\r}[1]{\frac1{x^{#1}}}$ 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?

Answer 1

Vandermonde's Identity yields $$ \begin{align} \sum_{k=0}^{20}\binom{30}{k}\binom{30}{k+10} &=\sum_{k=0}^{20}\binom{30}{k}\binom{30}{20-k}\\[6pt] &=\binom{60}{20} \end{align} $$

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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 $$ \sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}=(-1)^{15-\frac n2}\binom{30}{15-\frac n2} $$

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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 $$ \sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j-10}=\binom{30}{20} $$

Answer 2

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 $${30\choose 20}.$$