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(D) Let the set $S$ contain $2n + 1$ elements. We want to find the number of subsets containing more than $n$ elements,which means subsets with $n+1,

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$2^{2n}$

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(D) Let the set $S$ contain $2n + 1$ elements. We want to find the number of subsets containing more than $n$ elements,which means subsets with $n+1, n+2, \dots, 2n+1$ elements.The number of such subsets is given by the sum: $S = \binom{2n+1}{n+1} + \binom{2n+1}{n+2} + \dots + \binom{2n+1}{2n+1}$.Using the property of combinations $\binom{n}{r} = \binom{n}{n-r}$,we can rewrite the terms:$\binom{2n+1}{n+1} = \binom{2n+1}{n}$,$\binom{2n+1}{n+2} = \binom{2n+1}{n-1}$,and so on.Thus,$S = \binom{2n+1}{n} + \binom{2n+1}{n-1} + \dots + \binom{2n+1}{0}$.We know that the total number of subsets of a set with $2n+1$ elements is $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$.Since $\binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} = \binom{2n+1}{n+1} + \binom{2n+1}{n+2} + \dots + \binom{2n+1}{2n+1} = S$,we have $2S = 2^{2n+1}$.Therefore,$S = \frac{2^{2n+1}}{2} = 2^{2n}$.

$A$ set contains $2n + 1$ elements. The number of subsets of this set containing more than $n$ elements is equal to

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