Question:
The coefficient of $\mathrm{t}^{4}$ in the expansion of $\left(\frac{1-t^{6}}{1-t}\right)^{3}$ is
Correct Option: , 2
Solution:
$\left(1-t^{6}\right)^{3}(1-t)^{-3}$
$\left(1-t^{18}-3 t^{6}+3 t^{12}\right)(1-t)^{-3}$
$\Rightarrow$ cofficient of $t^{4}$ in $(1-t)^{-3}$ is
${ }^{3+4-1} C_{4}=6 C_{2}=15$
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