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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!

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    $\begingroup$ $22=26-4$, $21=25-4$, .. can you proceed from here ? $\endgroup$ – G Cab Oct 7 '16 at 21:00
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Hint: ${n \choose m}={n\choose n-m}$. Edit: Note that this should be easy to show both combinatorially and algebraically.

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  • $\begingroup$ I got $697585176,$ is that correct? $\endgroup$ – Dreamer Oct 7 '16 at 21:17
  • $\begingroup$ @Regina Using my hint, we can rewrite this expression so that each binomial coefficient has a coefficient of $9$. This is because for $i<13$, we have $(22-i){26 \choose i}$ and $(-4+i){26 \choose {26-i}}$ which we can sum to $(18){26 \choose i}$, and then redistribute it again into $9{26 \choose i}+9{26 \choose 26-i}$ by using the hint again. For $i=13$, we have only the term $9{26\choose 13}$. This gives us $9\sum_{i=0}^{26} {26\choose i}$. What does that sum represent combinatorially if you ignore the $9$? $\endgroup$ – Kevin Long Oct 7 '16 at 21:52
  • $\begingroup$ Is the answer $9 \cdot 2^{26}$? $\endgroup$ – Dreamer Oct 7 '16 at 22:00
  • $\begingroup$ @Regina That's right! $\endgroup$ – Kevin Long Oct 7 '16 at 22:13
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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{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}

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