Evaluation of a tricky binomial sum The Question:
$$
\mbox{To prove that:}\quad
\frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}
$$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$ as $\sum_{r=0}^{n}{(-1)^r\frac{n!3!}{(n-r)!(r+3)!}}$.
Now, since there is a term of $3!$ in it, I thought it would be a good idea to convert the term inside the expression to a $\binom{n+3}{r+3}$ term (by multiplying and dividing by $(n+1)(n+2)(n+3)$ i.e.
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\frac{(n+3)!}{(n-r)!(r+3)!}}$.
This becomes,
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\binom{n+3}{r+3}}$
Beyond this, I absolutely have no clue. I have worked at this for hours but I still can't seem to get an alternative to this method so, I tried sticking to it but unfortunately, I couldn't come up with anything. I just can't seem to figure out what I can do to further simplify this expression!!Any hint on how to progress will be greatly appreciated.
Thanks!
 A: Start from 
$$
\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}(-1)^r \binom{n+3}{r+3}
$$Reindex:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r-3} \binom{n+3}{r}
$$
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r} \binom{n+3}{r}
$$Now add and subtract the terms $0\leq r\leq 2$:
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\left(\sum_{r=0}^{n+3}(-1)^{r} \binom{n+3}{r}-\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right)
$$The binomial theorem graciously takes care of the first series for us, and then the result falls out:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\left(0+\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right)
$$
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\cdot \frac{(n+1)(n+2)}{2} = \frac{3}{n+3}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{r = 0}^{n}\pars{-1}^{r}\,{{n \choose r} \over {r + 3 \choose r}} & =
3\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r}
{\Gamma\pars{r + 1}\Gamma\pars{3} \over \Gamma\pars{r + 4}}
\\[5mm] & =
3\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r}
\int_{0}^{1}t^{r}\pars{1 - t}^{2}\,\dd t
\\[5mm] & =
3\int_{0}^{1}\bracks{\sum_{r = 0}^{n}{n \choose r}
\pars{-t}^{r}}\pars{1 - t}^{2}\,\dd t =
3\int_{0}^{1}\pars{1 - t}^{n}\pars{1 - t}^{2}\,\dd t
\\[5mm] & =\left. -3\,{\pars{1 - t}^{n + 3} \over n + 3}\right\vert_{0}^{1} =
{3 \over n + 3} = \bbx{3! \over 2\pars{n + 3}}
\end{align}
