A) 4
B) 8
C) 16
D) 32
Correct Answer: A
Solution :
Since the person is allowed to select at most n coins out of (2n + 1) coins, therefore in order to select one, two, three, ?., n coins. Thus, if T is the total number of ways of selecting one coin, then \[T={{\,}^{2n+1}}{{C}_{1}}\,{{+}^{2n+1}}{{C}_{2}}\,+\,......+{{\,}^{2n+1}}{{C}_{n}}\,=255\] ?..(i) Again the sum of binomial coefficients = \[^{2n+1}{{C}_{0}}\,{{+}^{2n+1}}{{C}_{1}}\,+{{\,}^{2n+1}}{{C}_{2}}+.....{{+}^{2n+1}}{{C}_{n}}\,{{+}^{2n+1}}{{C}_{n+1}}\]\[+{{\,}^{2n+1}}{{C}_{n+2}}\,+.....{{+}^{2n+1}}{{C}_{2n+1}}={{(1+1)}^{2n+1}}={{2}^{2n+1}}\] Þ \[^{2n+1}{{C}_{0}}\,+\,2\left( ^{2n+1}{{C}_{1}}\,+{{\,}^{2n+1}}{{C}_{2}}\,+...{{+}^{2n+1}}{{C}_{n}} \right)\]\[+{{\,}^{2n+1}}{{C}_{2n+1}}\,={{2}^{2n+1}}\] Þ \[1+2(T)+1={{2}^{2n+1}}\Rightarrow 1+T=\frac{{{2}^{2n+1}}}{2}={{2}^{2n}}\] Þ \[1+255={{2}^{2n}}\,\Rightarrow \,\,{{2}^{2n}}\,\,=\,\,{{2}^{8}}\,\Rightarrow \,\,n\,\,=\,\,4\].
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