Binomial series sum of the form $\sum^{k}_{r=0}(-1)^r(2)^{k-r}\binom{20}{r}\binom{20-r}{20-k}$ the value of 
$$2^k\binom{20}{0}\binom{20}{20-k}-2^{k-1}\binom{20}{1}\binom{19}{20-k}+2^{k-2}\binom{20}{2}\binom{18}{20-k} \cdots+ +(-1)^k\binom{20}{k}\binom{20-k}{20-k}$$
options: 
$(a)\;\; 7$
$(b)\;\;8$
$(c)\;\; 10$
$(d)\;\; 20$
Attempt: $$\sum^{k}_{r=0}(-1)^r(2)^{k-r}\binom{20}{r}\binom{20-r}{20-k} = (2)^{k}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\frac{20!}{r!\times (20-r)!}\cdot \frac{(20-r)!}{(20-k)!\times (k-r)!}$$
$$ = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\frac{k!}{r!\cdot (k-r)!} = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\binom{k}{r}$$
$$ = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\times \frac{1}{2^k} = \binom{20}{k}$$
none of the options is match.
please help me solve it, thanks
 A: The only possibility is D with k=1 or 19.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\sum_{r = 0}^{k}\pars{-1}^{r}\,\pars{2}^{k - r}{20 \choose r}
{20 - r \choose 20 - k} =
\bracks{z^{k}}\sum_{\ell = 0}^{\infty}z^{\ell}\bracks{2^{\ell}
\sum_{r = 0}^{\ell}\pars{-\,{1 \over 2}}^{r}{20 \choose r}
{20 - r \choose \ell - r}}
\\[5mm] = &\
\bracks{z^{k}}\sum_{r = 0}^{\infty}\pars{-\,{1 \over 2}}^{r}
{20 \choose r}\bracks{
\sum_{\ell = r}^{\infty}{20 - r \choose \ell - r}\pars{2z}^{\ell}}
\\[5mm] = &\
\bracks{z^{k}}\sum_{r = 0}^{\infty}\pars{-\,{1 \over 2}}^{r}
{20 \choose r}\bracks{
\sum_{\ell = 0}^{\infty}{20 - r \choose \ell}\pars{2z}^{\ell + r}}
\\[5mm] = &\
\bracks{z^{k}}\sum_{r = 0}^{\infty}\pars{-\,{1 \over 2}}^{r}
{20 \choose r}\pars{2z}^{r}\bracks{
\sum_{\ell = 0}^{\infty}{20 - r \choose \ell}\pars{2z}^{\ell}} =
\bracks{z^{k}}\sum_{r = 0}^{\infty}
{20 \choose r}\pars{-z}^{r}\pars{1 + 2z}^{20 - r}
\\[5mm] = &\
\bracks{z^{k}}\pars{1 + 2z}^{20}\sum_{r = 0}^{\infty}
{20 \choose r}\pars{-\,{z \over 1 + 2z}}^{r}
\\[5mm] = &\
\bracks{z^{k}}\pars{1 + 2z}^{20}\pars{1 - {z \over 1 + 2z}}^{20} =
\bracks{z^{k}}\pars{1 + z}^{20} = \bbx{\ds{20 \choose k}}
\end{align}
