Partial Sum of Alternating Sequence Involving Binomial Coefficients

Question

I am interested in a closed form solution for the sum: \begin{equation} \sum_{k=r}^m (-1)^{k-r}{q \choose k}{k \choose {k-r}}~. \end{equation} Entering this into Wolfram Alpha yields: \begin{equation} \frac{(-1)^{m-r+1}(m-r+1)}{r-q}{q \choose {m+1}}{{m+1} \choose {m-r+1}} \end{equation} Would love some help with proving this. Thanks!

Answer 1

Let $ q,r,m\in\mathbb{N} $, we have for any $ k\geq r : $

\begin{aligned}\binom{q}{k}\binom{k}{k-r}=\frac{q!}{\left(q-k\right)!r!\left(k-r\right)!}&=\frac{\left(q-r\right)!}{\left(k-r\right)!\left(q-k\right)!}\times\frac{q!}{r!\left(q-r\right)!}\\&=\binom{q-r}{k-r}\binom{q}{r}\end{aligned}

Thus :

\begin{aligned} \sum_{k=r}^{m}{(-1)^{k-r}\binom{q}{k}\binom{k}{k-r}}&=\binom{q}{r}\sum_{k=r}^{m}{\left(-1\right)^{k-r}\binom{q-r}{k-r}}\\ &=\binom{q}{r}\sum_{k=0}^{m-r}{\left(-1\right)^{k}\binom{q-r}{k}}\\ &=\binom{q}{r}+\binom{q}{r}\sum_{k=1}^{m-r}{\left(\left(-1\right)^{k}\binom{q-r-1}{k}-\left(-1\right)^{k-1}\binom{q-r-1}{k-1}\right)}\\ &=\binom{q}{r}+\binom{q}{r}\left(\left(-1\right)^{m-r}\binom{q-r-1}{m-r}-1\right)\\ &=\left(-1\right)^{m-r}\binom{q}{r}\binom{q-r-1}{m-r} \end{aligned}

(Note that in the third line we used Pascal's triangle formula : $ \binom{q-r}{k}=\binom{q-r-1}{k}+\binom{q-r-1}{k-1} $).