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$\ds{\sum_{i = 0}^{k}\pars{-1}^{k - i}{d - i \choose k - i}{n \choose i} =
{n - d + k - 1 \choose k}:\ {\large ?}.
\qquad\qquad 0 \leq 2k \leq d \leq n}$.
\begin{align}
\sum_{i = 0}^{k}\pars{-1}^{k - i}{d - i \choose k - i}{n \choose i} & =
\sum_{i = 0}^{k}\pars{-1}^{k - i}\braces{%
{-\bracks{d - i} + \bracks{k - i} - 1 \choose k - i}\pars{-1}^{k - i}}
{n \choose i}
\\[5mm] & =
\sum_{i = 0}^{k}{k - d - 1 \choose k - i}{n \choose i} =
\sum_{i = 0}^{k}{k - d - 1 \choose k - \bracks{k - i}}{n \choose k - i}
\\[5mm] & =
\sum_{i = 0}^{\color{#f00}{\infty}}{k - d - 1 \choose i}{n \choose k - i}
\qquad\pars{~\mbox{because}\
\left.{n \choose k - i}\right\vert_{\ i\ >\ k\,,\ n\ \geq\ 0} = 0~}
\\[5mm] & =
\sum_{i = 0}^{\infty}{k - d - 1 \choose i}
\bracks{z^{k - i}}\pars{1 + z}^{n}
\\[5mm] & =
\bracks{z^{k}}\bracks{\pars{1 + z}^{n}
\sum_{i = 0}^{\infty}{k - d - 1 \choose i}z^{i}}
\\[5mm] & =
\bracks{z^{k}}\bracks{\pars{1 + z}^{n}\pars{1 + z}^{k - d - 1}} =
\bracks{z^{k}}\pars{1 + z}^{n + k - d - 1}
\\[5mm] & =\
\bbox[#ffe,15px,border:1px dotted navy]{\ds{n + k - d - 1 \choose k}}
\end{align}