Explanation

(c)
General terms, $T_{r + 1} = (1)^{r}\ \ ^{15}C_{r}\$ $(x^{4})^{15 - r} \bullet \left( \frac{1}{x^{3}} \right)^{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = {( - 1)^{r}\ }^{15}C_{r}\$ ∙$x^{60 - 7r}$
For the coefficient of $x^{- 17},\$put $60 - 7r = - 17$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 60 + 17 = 7r \Longrightarrow r = 11$
Now, coefficient of $x^{- 17} = {( - 1)^{11}}^{15}C_{11}$ $= -^{15}C_{11}\$