Find the coefficient of $a^5{b}^7$ in $(a-2b{)}^{12}$

It is know 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 $a^5{b}^7$ occurs in the $(r+1{)}^{th}$ term of the expansion $(a-2b{)}^{12}$ we obtain 
$T_{r+1}{=}^{12}{C}_r(a{)}^{12-r}(-2b{)}^r{=}^{12}{C}_r(-2{)}^r(a{)}^{12-r}(b{)}^r$
Comparing the indices of a and b in $a^5{b}^7$ and in $T_{r+1}$, we obtain $r=7$
Thus the coefficient of $a^5{b}^7$ is
$={}^{12}{C}_7(-2{)}^7=\frac{12!}{7!5!}\times 2^7=\frac{\mathrm{12.11.10.9.8.7}!}{\mathrm{5.4.3.2.7}!}.2^7=-(792)(128)=-101376$