Question

The coefficient of $c^2{d}^2$ in the expansion of $(\sqrt{c}+\sqrt{d}{)}^8$ is ___

Answer 1

Answer: 70

$(\sqrt{c}+\sqrt{d}{)}^8,$ General term $={}^8{C}_r{c}^{\frac{\gamma -r}{2}}{d}^{\frac{r}{2}}$
For, $c^2{d}^2,r=4$
$\therefore$Coefficient $={}^8{C}_4=\frac{8\times 7\times 6\times 5}{24}=70$