15. Given a set $S$ satisfying $|S|=10$. Let $A_{1}, A_{2}, \cdots, A_{k}$ be non-empty subsets of $S$, and the intersection of any two subsets contains at most two elements. Find the maximum value of $k$.

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15. First, each $A_{i}(i=1,2, \cdots, k)$ contains at most 3 elements. Suppose $A_{j} \subseteq S, B$ has more than 3 elements. Let $a \in A_{j}$, since $A_{j} \cap\left(A_{j} \setminus \{a\}\right)$ has more than 2 elements, $A_{j} \setminus \{a\}$ is not one of the sets $A_{1}, A_{2}, \cdots, A_{k}$. Therefore, when $A_{j} \setminus \{a\}$ replaces $A_{j}$, the condition of the problem is still satisfied ($A_{j} \setminus \{a\}$ has one fewer element than $A_{j}$). Repeating this process, we can eventually ensure that the number of elements in the resulting set does not exceed 3, and the number of subsets that meet the condition does not increase after each replacement. All subsets of $S$ with at most 3 elements meet the condition, so the maximum value of $k$ is $C_{10}^{1} + C_{10}^{2} + C_{10}^{3} = 175$.