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, find the value of n.
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: In this question we use the theory of permutation and combination. Selecting at least one book means he can select 1 book , 2 books ,........or 2n+1 books. So, write the formula for selecting each of the number of books separately and add them to get 63.

Complete step-by-step answer:
So, before solving this question you need to first recall the basics of this chapter. For example, if we need to select two numbers out of four numbers. In this case, This can be done in${}^{\text{4}}{{\text{C}}_2}$ =6 ways.
Since the student is allowed to select at the most n-books out of (2n + 1) books,
Therefore, he can choose one book, two books or at the most n books.
And the number of ways of selecting at least one book are S (we say).
As we earlier mentioned,
  ${}^{2{\text{n + 1}}}{{\text{C}}_1} + {}^{2{\text{n + 1}}}{{\text{C}}_2} + ...... + {}^{2{\text{n + 1}}}{{\text{C}}_{\text{n}}} = 63$ =S(Say)
Again, we know that:
${}^{2{\text{n + 1}}}{{\text{C}}_0} + {}^{2{\text{n + 1}}}{{\text{C}}_1} + {}^{2{\text{n + 1}}}{{\text{C}}_2} + ...... + {}^{2{\text{n + 1}}}{{\text{C}}_{\text{n}}} + ... + {}^{2{\text{n + 1}}}{{\text{C}}_{2{\text{n + 1}}}} = {2^{2{\text{n + 1}}}}$

Now,
${}^{2{\text{n + 1}}}{{\text{C}}_0} = {}^{2{\text{n + 1}}}{{\text{C}}_{2n + 1}} = 1$
And ${}^{2{\text{n + 1}}}{{\text{C}}_1} = {}^{2{\text{n + 1}}}{{\text{C}}_{2n}}$ and so on

Hence, we have
        1 + 1 + 2S = ${2^{2{\text{n + 1}}}}$
   $ \Rightarrow $ 2 + 2 x 63 = ${2^{2{\text{n + 1}}}}$
 $ \Rightarrow $ 128 = ${2^{2{\text{n + 1}}}}$ or ${2^7}$ = ${2^{2{\text{n + 1}}}}$
 $ \Rightarrow $ ${2^{2{\text{n + 1}}}}$ = 7
    $ \Rightarrow $ 2n = 6
     $ \Rightarrow $ n = 3

Therefore, the value of n is 3.
And so, option (C) is correct.

Note: In this question, we have to do selection. So, we will use the concept of combination not permutation. We need to remember this formula for selecting r things out of n.
                                ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
For example, if we want to choose 1 book out of 3 books. This can be done in \[{}^3{{\text{C}}_1}\] =3 ways.