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

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(D) Let $S$ be the number of subsets containing at most $n$ elements. Then,$S = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n}$.We know t

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$2^{2n}$

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(D) Let $S$ be the number of subsets containing at most $n$ elements. Then,$S = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n}$.We know that the total number of subsets of a set with $(2n+1)$ elements is $2^{2n+1}$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have:$\binom{2n+1}{0} = \binom{2n+1}{2n+1}$$\binom{2n+1}{1} = \binom{2n+1}{2n}$$\dots$$\binom{2n+1}{n} = \binom{2n+1}{n+1}$Summing these,we get $2S = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.Therefore,$S = \frac{2^{2n+1}}{2} = 2^{2n}$.

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

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