Evaluate the sum $ 22\binom{26}{0} + 21\binom{26}{1} + 20\binom{26}{2} + \cdots + (-3)\binom{26}{25} + (-4)\binom{26}{26}. $ Evaluate the sum
$$22\binom{26}{0} + 21\binom{26}{1} + 20\binom{26}{2} + \cdots + (-3)\binom{26}{25} + (-4)\binom{26}{26}.$$

Is there an obvious shortcut I'm missing?  I know I can't just do the whole calculation!  That would take years!  But I don't see another way.  Solutions are greatly appreciated!
 A: Hint: ${n \choose m}={n\choose n-m}$.
Edit: Note that this should be easy to show both combinatorially and algebraically.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \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}}}
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\begin{align}
\sum_{k = 0}^{26}\pars{22 - k}{26 \choose k} & =
22\sum_{k = 0}^{26}{26 \choose k} -
\left.\partiald{}{x}\sum_{k = 0}^{26}{26 \choose k}x^{k}\,\right\vert_{\ x\ =\ 1}
=
22 \times 2^{26} - \left.\partiald{\pars{1 + x}^{26}}{x}\,\right\vert_{\ x\ =\ 1}
\\[5mm] & =
22 \times 2^{26} - 26 \times 2^{25} = 22 \times 2^{26} - 13 \times 2^{26} =
\color{#f00}{9 \times 2^{26}}
\end{align}
