Question

A person is permitted to select at least one and at most $n$ coins from a collection of $(2n+1)$ distinct coins. If the total number of ways in which he can select coins is $255$, then $n$ equals

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

Answer: (A)

$4$

We have
${}^{2n+1}{C}_1{+}^{2n+1}{C}_2+\cdots {+}^{2n+1}{C}_n=255$ ...(i)
Also, the sum of binomial coefficients
${}^{2n+1}{C}_0{+}^{2n+1}{C}_1{+}^{2n+1}{C}_2+\cdots {+}^{2n+1}{C}_n{+}^{2n+1}{C}_{n+1}{+}^{2n+1}{C}_{n+2}+\cdots {+}^{2n+1}{C}_{2n+1}={(1+1)}^{2n+1}=2^{2n+1}$
${\Rightarrow }^{2n+1}{C}_0+2\{{}^{2n+1}{C}_1{+}^{2n+1}{C}_2+\cdots {+}^{2n+1}{C}_n\}{+}^{2n+1}{C}_{2n+1}=2^{2n+1}$
$\Rightarrow 1+2\left(255\right)+1=2^{2n+1}$
$\Rightarrow 1+255=2^{2n}$
$\Rightarrow 2^{2n}=2^8$
$\Rightarrow n=4$