A) 2
B) 3
C) 4
D) None of these
Correct Answer: B
Solution :
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 the choice to select one, two, three, ......, n books. Thus, if T is the total number of ways of 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 \[^{2n+1}{{C}_{0}}+{{\,}^{2n+1}}{{C}_{1}}+{{\,}^{2n+1}}{{C}_{2}}+.....+{{\,}^{2n+1}}{{C}_{n}}+{{\,}^{2n+1}}{{C}_{n+1}}\]\[{{+}^{2n+1}}{{C}_{n+2}}+....+{{\,}^{2n+1}}{{C}_{2n+1}}={{(1+1)}^{2n+1}}={{2}^{2n+1}}\] or \[^{2n+1}{{C}_{0}}+2{{(}^{2n+1}}{{C}_{1}}+{{\,}^{2n+1}}{{C}_{2}}+..+{{\,}^{2n+1}}{{C}_{n}}){{+}^{2n+1}}{{C}_{2n+1}}={{2}^{2n+1}}\] Þ \[1+2(T)+1={{2}^{2n+1}}\] Þ \[1+T=\frac{{{2}^{2n+1}}}{2}={{2}^{2n}}\] Þ \[1+63={{2}^{2n}}\Rightarrow {{2}^{6}}={{2}^{2n}}\Rightarrow n=3\].You need to login to perform this action.
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