Explanation

(b)
$\frac{1}{\ ^{4}C_{n}} = \frac{1}{\ ^{5}C_{n}} + \frac{1}{\ ^{6}C_{n}}$
$\Rightarrow \ \ \ \frac{n!(4 - n)!}{4!} = \frac{n!(5 - n)!}{5!} + \frac{n!(6 - n)!}{6!}$
$\Rightarrow \ \ \ \frac{(4 - n)!}{4!} = \frac{(4 - n)!(5 - n)}{5 \times 4!} + \frac{(6 - n)(5 - n)(4 - n)!}{6 \times 5 \times 4!}$
$\Rightarrow \ \ \ 1 = \frac{5 - n}{5} + \frac{(6 - n)(5 - n)}{6 \times 5}$
$\Rightarrow \ \ n^{2} - 17n + 30 = 0$
$\Rightarrow \ \ \ (n - 15)(n - 2) = 0$
$\Rightarrow \ \ \ \ n = 2$
[$\because\ \ ^{4}C_{n}\$ is not meaningful for $n = 15$]