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(C) Let $S_n$ be the number of subsets $A \subseteq \{1, 2, \ldots, n\}$ such that any two elements differ by at least $3$. Let $a_n$ be the number of

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$p > q$ and $p - q = \frac{1}{10}$

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(C) Let $S_n$ be the number of subsets $A \subseteq \{1, 2, \ldots, n\}$ such that any two elements differ by at least $3$. Let $a_n$ be the number of such subsets. For $n=10$,we calculate the total number of valid subsets $N$. Let $f(n)$ be the number of such subsets for $X_n$. The recurrence relation is $f(n) = f(n-1) + f(n-3) + 1$ (where $1$ accounts for the empty set). Calculating values: $f(0)=1, f(1)=2, f(2)=3, f(3)=4, f(4)=6, f(5)=9, f(6)=13, f(7)=19, f(8)=28, f(9)=41, f(10)=60$. Total subsets $N = 60$. Number of subsets containing $1$: If $1 \in A$,then $2, 3 
otin A$. We need to choose a subset from $\{4, 5, \ldots, 10\}$ such that elements differ by at least $3$. This is equivalent to choosing a subset from $\{1, 2, \ldots, 7\}$ with the same condition. Thus,$N(1 \in A) = f(7) = 19$. So,$p = \frac{19}{60}$. Number of subsets containing $2$: If $2 \in A$,then $1, 3, 4 
otin A$. We need to choose a subset from $\{5, 6, \ldots, 10\}$ such that elements differ by at least $3$. This is equivalent to choosing a subset from $\{1, 2, \ldots, 6\}$ with the same condition. Thus,$N(2 \in A) = f(6) = 13$. So,$q = \frac{13}{60}$. Therefore,$p > q$ and $p - q = \frac{19-13}{60} = \frac{6}{60} = \frac{1}{10}$.

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