If a set A has 2n + 1 elements,then how many | vedclass

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(C) Let $S$ be the set $A$ with $|A| = 2n + 1$. We want to find the number of subsets with at least $n$ elements. This is given by the sum: $\sum_{k=n

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$2^{2n+1}$

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(C) Let $S$ be the set $A$ with $|A| = 2n + 1$. We want to find the number of subsets with at least $n$ elements. This is given by the sum: $\sum_{k=n}^{2n+1} \binom{2n+1}{k}$.We know that the total number of subsets is $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$.Let $X = \sum_{k=n}^{2n+1} \binom{2n+1}{k}$.Using the property $\binom{m}{r} = \binom{m}{m-r}$,we have $\sum_{k=0}^{n-1} \binom{2n+1}{k} = \sum_{k=n+2}^{2n+1} \binom{2n+1}{k}$.Also,$\sum_{k=0}^{2n+1} \binom{2n+1}{k} = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.Since $\binom{2n+1}{n} = \binom{2n+1}{n+1}$,the sum of the first $n+1$ terms is equal to the sum of the last $n+1$ terms.Let $S_1 = \sum_{k=0}^{n} \binom{2n+1}{k}$ and $S_2 = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k}$.Since $S_1 + S_2 = 2^{2n+1}$ and $S_1 = S_2$ (by symmetry),we have $S_2 = \frac{2^{2n+1}}{2} = 2^{2n}$.Thus,the number of subsets with at least $n+1$ elements is $2^{2n}$.The question asks for at least $n$ elements,which is $\binom{2n+1}{n} + S_2 = \binom{2n+1}{n} + 2^{2n}$.

If a set $A$ has $2n + 1$ elements,then how many subsets of $A$ contain at least $n$ elements?

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