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

Binomial Theorem

Solution:

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