<s>[INST] 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
 [/INST] The only possibility is D with k=1 or 19.
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