Explanation

(a)
$\frac{2}{9!} + \frac{2}{3!7!} + \frac{1}{5!5!}$
$= \frac{1}{1!9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{3!7!} + \frac{1}{9!1!}$
$= \frac{1}{10!}\left\lbrack \frac{10!}{1!9!} + \frac{10!}{3!7!} + \frac{10!}{5!5!} + \frac{10!}{3!7!} + \frac{10!}{9!1!} \right\rbrack$
$= \frac{1}{10!}\{\ ^{10}C_{1} + \ ^{10}C_{3} + \ ^{10}C_{5} + \ ^{10}C_{7} + \ ^{10}C_{9}\}$
$= \frac{1}{10!}\left( 2^{10 - 1} \right) = \frac{2^{9}}{10!} = \frac{2^{a}}{b!}$ (given)
$\Rightarrow \ a = 9,\ b = 10$