# Find the coefficient of a5b7 in (a – 2b)12

Question:

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

Solution:

It is known that $(r+1)^{\text {th }}$ term,$\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $\mathrm{T}_{\mathrm{r}+1}={ }^{n} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$.

Assuming that $a^{5} b^{7}$ occurs in the $(r+1)^{\text {th }}$ term of the expansion $(a-2 b)^{12}$, we obtain

$\mathrm{T}_{r+1}={ }^{12} \mathrm{C}_{r}(\mathrm{a})^{12-r}(-2 \mathrm{~b})^{r}={ }^{12} \mathrm{C}_{\mathrm{r}}(-2)^{r}(\mathrm{a})^{12-r}(\mathrm{~b})^{r}$

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 ${ }^{12} C_{7}(-2)^{7}=-\frac{12 !}{7 ! 5 !} \cdot 2^{7}=-\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8.7 !}{5 \cdot 4 \cdot 3 \cdot 2.7 !} \cdot 2^{7}=-(792)(128)=-101376$