Explanation

(a)
Since, the person is allowed to select at most $n$ coins out of $(2n + 1)$ coins, therefore in order to select one, two, three,…., $n$ coins, if $T$ is the total number of ways of selecting at least one coin, then
$T = {\ ^{2n + 1}C}_{1} +^{2n + 1}C_{2} + ... +^{2n + 1}C_{n} = 255$ …(i)
Using the binomial theorem
$\ ^{2n + 1}C_{0} +^{2n + 1}C_{1} +^{2n + 1}C_{2} + ... +^{2n + 1}C_{n}$
$+^{2n + 1}C_{n + 1} +^{2n + 1}C_{n + 2} + ... +^{2n + 1}C_{2n + 1}$
$= (1 + 1)^{2n + 1} = 2^{2n + 1}$
${\Rightarrow \ }^{2n + 1}C_{0} + 2\left( \ ^{2n + 1}C_{1} +^{2n + 1}C_{2} \right.\ + ....$
$+^{2n + 1}C_{n}) +^{2n + 1}C_{2n + 1} = 2^{2n + 1}$
$\Rightarrow 1 + 2(T) + 1 = 2^{2n + 1}$
$\Rightarrow 1 + T = \frac{2^{2n + 1}}{2} = 2^{2n}$ [from Eq.(i)]
$\Rightarrow 1 + 255 = 2^{2n}$
$\Rightarrow 2^{2n} = 2^{8} \Rightarrow n = 4$