Explanation

(b)
On putting $x = - 1$in
${(1 + x)}^{20} = \ ^{20}C_{0} + \ ^{20}C_{1}x + ... + \ ^{20}C_{10}x^{10} + \ldots + \ ^{20}C_{10}x^{10},$
we get
$0 = \ ^{20}C_{0} - \ ^{20}C_{1} + \ldots - \ ^{20}C_{9} + \ ^{20}C_{10} - \ ^{20}C_{11} + \ldots + \ ^{20}C_{20}$
$\Longrightarrow 0 = \ ^{20}C_{0} - \ ^{20}C_{1} + \ldots - \ ^{20}C_{9} + \ ^{20}C_{10} - \ ^{20}C_{9} + \ldots + \ ^{20}C_{0}$
$\Longrightarrow 0 = 2\left( \ ^{20}C_{0} - \ ^{20}C_{1} + \ldots - \ ^{20}C_{9} \right) + \ ^{20}C_{10}$
$\Longrightarrow \ \ ^{20}C_{10} = 2\left( \ ^{20}C_{0} - \ ^{20}C_{1} + \ldots + \ ^{20}C_{10} \right)$
$\Longrightarrow \ ^{20}C_{0} - \ ^{20}C_{1} + ... + \ ^{20}C_{10} = \frac{1}{2}\ ^{20}C_{10}$