Question

A student is allowed to select at most n books from a collection of $ p(x)=a{{x}^{2}}+bx+c $ books. If the total number of ways in which he can select one book is 63, then the value of n is equal to :

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

Answer: (B)

Since the student is allowed to select at most n books out of $ (2n+1) $ books, $ \therefore $ In order to select one book he has to select one book he has the choice to select one, two, three ... n books, thus if T is the total number of ways selecting one book then $ T{{=}^{2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+.....{{+}^{2n+1}}{{C}_{n}}=63 $ ?(i) again the sum of binomial coefficients $ ^{2n+1}{{C}_{0}}{{+}^{2n+1}}{{C}_{2}}+.....{{+}^{2n+1}}{{C}_{n}}{{+}^{2n+1}}{{C}_{n+1}}+... $ $ ={{(1+1)}^{2n+1}}={{2}^{2n+1}} $ Or $ ^{2b+1}{{C}_{0}}+2\left( ^{2n+1}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+....{{+}^{2n+1}}{{C}_{n}} \right) $ $ {{+}^{2n+1}}{{C}_{2n+1}}={{2}^{2n+1}} $ $ \Rightarrow $ $ 1+2(T0+1={{2}^{2n+1}} $ $ \Rightarrow $ $ 1+T=\frac{{{2}^{2n+1}}}{2}={{2}^{2n}} $ $ \Rightarrow $ $ 1+63={{2}^{2n}} $ $ \Rightarrow $ $ {{2}^{6}}={{2}^{2n}} $ $ \Rightarrow $ $ n=3 $