3

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

DXT
  • 11,241

2 Answers2

3

The only possibility is D with k=1 or 19.

marty cohen
  • 107,799
2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
  • 89,464
  • 1
    Isn't this reproving an identity already proved in the question? – Did Apr 09 '17 at 17:18
  • @Did I suspect that at the very beginning. I'll wait 24 four hours to delete it such that you can read this comment. Thanks. – Felix Marin Apr 09 '17 at 22:03
  • Please don't delete this answer because it proves a identity which the OP used without any proof. – Navin Apr 10 '17 at 10:58
  • @navinstudent You're right. Basically, my answer is related to a $\texttt{Generating Function}$ while the OP performs a 'disentangling' of the binomial coefficients. Thanks. – Felix Marin Apr 10 '17 at 20:11