Find the coefficient of $x^5$ in $(x+3{)}^8$

It is known that $(r+1)$th term $(T_{r+1})$ in the binomial expansion of
$(a+b{)}^n$ is given by $T_{r+1}{=}^n{C}_r{a}^{a-r}{b}^r$
Assuming that $x^5$ occurs in the $(r+1)$th term of expansion $(x+3{)}^8$ we obtain 
 $T_{r+1}{=}^8{C}_r(x{)}^{8-r}(3{)}^r$
Comparing the indices of $x$ in $x^5$ and in $T_{r+1}$, we obtain $r=3$
Thus, the coefficient of $x^5$ is 
${}^8{C}_3(3{)}^3=\frac{8!}{3!5!}x{3}^3=\frac{\mathrm{8.7.6.5}!}{\mathrm{3.2.5}!}.3^3=1512$