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 he can select at least one book is $63$, then the value of n is equal to

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct answer:

Step 1: Finding the number of ways of selection.

A student is allowed to select at most $n$ books out of $(2n+1)$ books.

Number of ways, he can select at least one book = $63$

This selection is given by the combination concepts,

${}^{2n+1}C_1+{}^{2n+1}C_2+{}^{2n+1}C_3+........+{}^{2n+1}C_n=63...\left(1\right)$

we know that, ${}^{n}C_r={}^{n}C_{n-r}$. and

${}^{2n+1}C_0+{}^{2n+1}C_1+{}^{2n+1}C_2+........+{}^{2n+1}C_n=2^{2n+1}$.

Step 2: Finding the value of n.

Thus the equation becomes,

$$\begin{gathered} \Rightarrow 2({}^{2n+1}C_0+{}^{2n+1}C_1+{}^{2n+1}C_2+........+{}^{2n+1}C_n)=2^{2n+1} \\ \Rightarrow {}^{2n+1}C_0+{}^{2n+1}C_1+{}^{2n+1}C_2+........+{}^{2n+1}C_n=2^{2n} \\ \Rightarrow {}^{2n+1}C_1+{}^{2n+1}C_2+{}^{2n+1}C_3+........+{}^{2n+1}C_n=2^{2n}-1...\left[\because {}^{2n+1}C_0=1\right] \end{gathered}$$

On comparing equation $\left(1\right)$ with this equation,

$$\begin{gathered} \Rightarrow 2^{2n}-1=63 \\ \Rightarrow 2^{2n}=64 \\ \Rightarrow 2^{2n}=2^6 \\ \Rightarrow 2n=6 \\ \Rightarrow n=3 \end{gathered}$$

Hence, option (D) $\mathbf{3}$ is the correct answer.