Explanation

(b)
$General\ term,\ T_{r + 1} = \ ^{15}C_{r}(x^{4})^{15 - r}\left( - \frac{1}{x^{3}} \right)^{r}$
$= \ ^{15}C_{r}x^{60 - 7r}( - 1)^{r}$
For the coefficient of $x^{3r}$ put
$\ \ \ \ \ \ \ \ \ 60 - 7r = 32 \Longrightarrow r = 4$
Now, coefficient$\ of\ x^{32}\ in\left( x^{4} - \frac{1}{x^{3}} \right)^{15} = \ ^{15}C_{4}{( - 1)}^{4} = \ ^{15}C_{4}$