A) \[{{2}^{n}}\]
B) \[{{2}^{n+1}}\]
C) \[{{2}^{n-1}}\]
D) \[{{2}^{2n}}\]
Correct Answer: D
Solution :
The number of sub-sets of the set which contain at most n elements is \[^{2n+1}{{C}_{0}}{{+}^{2n+1}}{{C}_{1}}+.....+{{\,}^{2n+1}}{{C}_{n}}=S\] (Say) Then \[2S=2{{(}^{2n+1}}{{C}_{0}}+{{\,}^{2n+1}}{{C}_{1}}+.....+{{\,}^{2n+1}}{{C}_{n}})\] = \[{{(}^{2n+1}}{{C}_{0}}+{{\,}^{2n+1}}{{C}_{2n+1}})+{{(}^{2n+1}}{{C}_{1}}+{{\,}^{2n+1}}{{C}_{2n}})+........+{{(}^{2n+1}}{{C}_{n}}+{{\,}^{2n+1}}{{C}_{n+1}})\]\[\left\{ \because \,{{\,}^{n}}{{C}_{r}}={{\,}^{n}}{{C}_{n-r}} \right\}\] = \[^{2n+1}{{C}_{0}}+{{\,}^{2n+1}}{{C}_{1}}+.......+{{\,}^{2n+1}}{{C}_{2n+1}}={{2}^{2n+1}}\] \[\Rightarrow \] \[S={{2}^{2n}}\].
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