Question

A student is allowed to select at most $n$ books from a collection of $(2n + 1)$ books. If the total number of ways in which be can select a book is $255$, then the value of $n$ equals to

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

Answer: (C)

By the given condition
$^{2n+1}C_{1}+\,{}^{2n+1}C_{2}+ ..........+\,{}^{2n+1}C_{4}=255$
Now $^{2n+1}C_{1}=\,{}^{2n+1}C_{2n} \, \left[\because \,{}^{n}C_{r}=\,{}^{n}C_{n-r}\right]$
$^{2n+1}C_{2}=\,{}^{2n+1}C_{2n-1}$
$^{2n+1}C_{n} =\,{}^{2n+1}C_{n+1}$
Adding these, we get
$^{2n+1}C_{0}+\,{}^{2n+1}C_{1}+\,{}^{2n+1}C_{2}+...........+\,{}^{2n+1}C_{n}$
$=^{2n+1}C_{n+1}+\,{}^{2n+1}C_{2n}+\,{}^{2n+1}C_{2n+1}$
$\Rightarrow 2\left[^{2n+1}C_{0}+\,{}^{2n+1}C_{1}+..........+\,{}^{2n+1}C_{n}\right]$
$=2^{2n+1}$
$\Rightarrow \,{}^{2n+1}C_{0}+\,{}^{2n+1}C_{1}+ ..........+\,{}^{2n+1}C_{n}=2^{2n}$
$\Rightarrow \,{}^{2n+1}C_{1}+.........+^{2n+1}C_{n}=2^{2n}-1$
$\Rightarrow 255=2^{2n}-1$
$\Rightarrow 256=2^{n}$
$\Rightarrow 2^{8}=2^{2n}$
$\Rightarrow 2n=8$
$\Rightarrow n=4$.
$[\because$ $\displaystyle \sum_{r=0}^n $ $\,{}^{n}C_{r}=2^{n}]$