What is the value of sum S given by the following ?

$S=\sum _{r=0}^n{(-1)}^r\left(\frac{{ }^n{C}_r}{{ }^{r+3}{C}_r}\right)$

$$\begin{gathered} \frac{{ }^n{C}_r}{{ }^{r+3}{C}_r}=\frac{n!}{r!(n-r)!}\frac{r!3!}{(r+3)!} \\ =\frac{6.n!}{(n-r)!(r+3)!}=\frac{6}{(n+1)(n+2)(n+3)}.\frac{(n+3)!}{(r+3)!(n-r)!} \\ S=\frac{6}{(n+1)(n+2)(n+3)}\sum _{r=0}^n{(-1)}^r{}^{n+3}{C}_{r+3} \end{gathered}$$

If we represents the summation expression  in  (1) by $S_1$, then 
$S_1=\sum _{r=0}^n{{(-1)}^r}^{{}^{n+3}}{C}_{r+3}=\sum _{s=3}^{n+3}{(-1)}^{s-3n+3}{C}_s=-\sum _{s=3}^{n+3}{(-1)}^{sn+3}{C}_s$

$=-\{\sum _{s=0}^{n+3}(-1{)}^s{}^{n+3}{C}_s-{(}^{n+3}{C}_0{-}^{n+3}{C}_1{+}^{n+3}{C}_2\}$
The  term  $T_1$  is 0(why), while 
$T_2=1-(n+3)+\frac{(n+3)(n+2)}{2}=\frac{(n+1)(n+2)}{2}$
From (1) 

$s=\frac{3}{n+3}$