A set contains 2n + 1 elements. The number of | vedclass

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

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$2^{2n}$

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(D) Let the set $S$ contain $N = 2n + 1$ elements.We need 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 binomial coefficients $\binom{n}{r} = \binom{n}{n-r}$,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 sum of all binomial coefficients for a set of size $m$ is $2^m$,i.e.,$\sum_{k=0}^{m} \binom{m}{k} = 2^m$.For $m = 2n + 1$,the total sum is $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$.Since $\binom{2n+1}{k} = \binom{2n+1}{2n+1-k}$,the sum of the first half of the coefficients is equal to the sum of the second half.Therefore,$S = \frac{1}{2} 	imes 2^{2n+1} = 2^{2n}$.

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

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