Question

A Student is allowed to select at most, n book from a collection of (2n+1) books. If the total number of ways in which he can select a book is 63, then n Is equal
$
  {\text{A}}{\text{. 3}} \\
  {\text{B}}{\text{. 4}} \\
  {\text{C}}{\text{. 5}} \\
  {\text{D}}{\text{. 6}} \\
 $

Answer 1

Hint: In order to solve this problem we have to first assume that the number of ways of selecting at least one book be ${\text{S}}$. Using properties of Binomial Expansion like $_{}^n{C_r} = _{}^n{C_{n - r}}$ we resolve questions into some forms and then solve them.

Complete step-by-step answer:
Since a student is allowed to select at most n books from a combination of (2n+1) books. Therefore, he can choose one book, two books or at most n books.
The number of ways of selecting at least one book is
 \[ \Rightarrow {\text{ }}_{}^{2n + 1}{C_1} + _{}^{2n + 1}{C_2} + .............. + _{}^{2n + 1}{C_n} = 63{\text{ }}\]
Let above is equal to ${\text{S}}$, then
\[ \Rightarrow {\text{S }} = {\text{ }}_{}^{2n + 1}{C_1} + _{}^{2n + 1}{C_2} + .............. + _{}^{2n + 1}{C_n} = 63{\text{ }}\] eq.1
From Binomial expansion, we know that
\[{(1 + x)^n} = _{}^n{C_0} + _{}^n{C_1}{x^1} + ........... + _{}^nC_n^{}{x^n}\]
 on putting x = 1 in given equation we get
${2^n}{ = ^n}{C_0}{ + ^n}{C_1} + \cdots \cdots { + ^n}{C_n}$ eq.2
Similarly from binomial expansion for given question is result into
$ \Rightarrow _{}^{2n + 1}{C_0} + _{}^{2n + 1}{C_1} + .......... + _{}^{2n + 1}{C_{2n + 1}} = {2^{2n + 1}}{\text{ }}$ Eq.3
Now , from the properties of binomial expansion
${ \Rightarrow ^{2n + 1}}{C_r}{ = ^{2n + 1}}{C_{2n + 1 - r}}$ ( r is items taken at that time )
On putting r = 0 , we get,
$^{2n + 1}{C_0}{ = ^{2n + 1}}{C_{2n + 1}} = 1$


Now, we can rewrite eq.3 as
\[ \Rightarrow 1 + _{}^{2n + 1}{C_1} + ............ + _{}^{2n + 1}{C_n} + 1 = {2^{2n + 1}}\]
Using eq.1 we can rewrite above equation as
$ \Rightarrow 1 + 2{\text{S + 1 = }}{{\text{2}}^{2n + 1}}$
On further solving we get,
$ \Rightarrow 2 + 2{\text{S = }}{{\text{2}}^{2n + 1}}$
Since S = 63 then
$ \Rightarrow 2 + 2 \times 63 = {2^{2n + 1}}$
$ \Rightarrow 128 = {2^{2n + 1}}$
$
   \Rightarrow {2^7} = {2^{2n + 1}} \\
   \Rightarrow 7 = 2n + 1 \\
   \Rightarrow n = 3 \\
 $
Hence, option A is correct.

Note:
Whenever you get this type of question the key concept of solving is you have to convert question language into Permutation and Combination formulas and then using properties like summation of binomial coefficient and binomial expansion should be remembered to solve this type of questions.