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!

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:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\left(0+\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right)
$$
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\cdot \frac{(n+1)(n+2)}{2} = \frac{3}{n+3}
$$
