# The coefficient of

Question:

The coefficient of $a^{-6} b^{4}$ in the expansion $\left(\frac{1}{a}-\frac{2}{3} b\right)^{10}$ is

Solution:

$\operatorname{In}\left(\frac{1}{a}-\frac{2}{3} b\right)^{10}$'

$T_{r+1}={ }^{10} C_{r}\left(\frac{1}{a}\right)^{10-r}\left(\frac{-2 b}{3}\right)^{r}$

for coefficient of $a^{-6} b^{4}$

put $10-r=6$

i. e. $r=4$

We get

coefficient of $a^{-6} b^{4}={ }^{10} C_{4}\left(\frac{-2}{3}\right)^{4}$

$=\frac{10 !}{4 ! 6 !} \times \frac{2^{4}}{3^{4}}$

$=\frac{10 \times 9 \times 8 \times 7}{2 \times 3 \times 4} \times \frac{2^{4}}{3^{4}}$

$=\frac{10 \times 7 \times 2^{4}}{3^{3}}$

i. e. coefficient of $a^{-6} \quad b^{4}=\frac{1120}{27}$.