For n geq 2,let S_n denote the set of all sub | vedclass

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(A) The number of subsets of $\{1, 2, \ldots, n\}$ with $r$ elements such that no two elements are consecutive is given by the formula $\binom{n-r+1}{

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$13$

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(A) The number of subsets of $\{1, 2, \ldots, n\}$ with $r$ elements such that no two elements are consecutive is given by the formula $\binom{n-r+1}{r}$.For $n=5$,we calculate the number of such subsets for each possible size $r$:- For $r=0$ (empty set): $\binom{5-0+1}{0} = \binom{6}{0} = 1$.- For $r=1$: $\binom{5-1+1}{1} = \binom{5}{1} = 5$.- For $r=2$: $\binom{5-2+1}{2} = \binom{4}{2} = 6$.- For $r=3$: $\binom{5-3+1}{3} = \binom{3}{3} = 1$.Total number of subsets $n(S_5) = 1 + 5 + 6 + 1 = 13$.

For $n \geq 2$,let $S_n$ denote the set of all subsets of $\{1, 2, \ldots, n\}$ such that no two elements are consecutive. For example,$\{1, 3, 5\} \in S_6$,but $\{1, 2, 4\} \notin S_6$. Then $n(S_5)$ is equal to . . . . . . .

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