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

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(B) Let $S$ be a set with $N = 2n + 1$ elements. We want to find the number of subsets with at most $n$ elements,which is given by the sum: $S = \sum_

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$2^{2n}$

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(B) Let $S$ be a set with $N = 2n + 1$ elements. We want to find the number of subsets with at most $n$ elements,which is given by the sum: $S = \sum_{r=0}^{n} \binom{2n+1}{r} = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n}$. We know that the total number of subsets is $\sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$. By the property of combinations,$\binom{N}{r} = \binom{N}{N-r}$. Thus,$\binom{2n+1}{0} = \binom{2n+1}{2n+1}$,$\binom{2n+1}{1} = \binom{2n+1}{2n}$,...,$\binom{2n+1}{n} = \binom{2n+1}{n+1}$. Let $X = \sum_{r=0}^{n} \binom{2n+1}{r}$. Then $X = \binom{2n+1}{0} + \dots + \binom{2n+1}{n}$. Also,$X = \binom{2n+1}{2n+1} + \dots + \binom{2n+1}{n+1}$. Adding these two expressions: $2X = \sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$. Therefore,$X = \frac{2^{2n+1}}{2} = 2^{2n}$.

$A$ set contains $(2n + 1)$ elements. The number of subsets of the set which contain at most $n$ elements is:

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