$11 \cdot 94$ Let $S=\{1,2, \cdots, 15\}$, and from $S$ take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$, satisfying the following conditions:
(i) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(ii) $\left|A_{i} \cap A_{j}\right| \leqslant 3,1 \leqslant i<j \leqslant n$;
(iii) for any 3-element subset $M$ of $S$, there exists some $A_{k}, 1 \leqslant k \leqslant n$, such that $M \subset A_{k}$.
Find the minimum value of the number of such subsets $n$.
[Solution] If there is $a \in S$ that belongs to at most 6 subsets $A_{i_{1}}, A_{i_{2}}, \cdots A_{i_{6}}$, then each $A_{i_{j}}$ contains 6 other elements besides $a$, forming a total of $C_{6}^{2}=15$ different triples containing $a$. Thus, the 6 subsets can form at most 90 different triples containing $a$.

On the other hand, the total number of different triples containing $a$ in $S$ is $C_{14}^{2}=7 \times 13=91$ $>90$, making it impossible for (iii) to hold. Therefore, to satisfy conditions (i)-(iii), each element in $S$ must belong to at least 7 subsets. This implies that $n \geqslant 15$.
Using lexicographic order, we can write 15 7-element subsets that meet the requirements of the problem as follows:
$$
\begin{array}{ll}
\{1,2,3,4,5,6,7\}, & \{1,2,3,8,9,10,11\}, \\
\{1,2,3,12,13,14,15\}, & \{1,4,5,8,9,12,13\}, \\
\{1,4,5,10,11,14,15\}, & \{1,6,7,8,9,14,15\}, \\
\{1,6,7,10,11,12,13\}, & \{2,4,6,8,10,12,14\}, \\
\{2,4,6,9,11,13,15\}, & \{2,5,7,8,10,13,15\}, \\
\{2,5,7,9,11,12,14\}, & \{3,4,7,8,11,12,15\}, \\
\{3,4,7,9,10,13,14\}, & \{3,5,6,8,11,13,14\}, \\
\{3,5,6,9,10,12,15\} . &
\end{array}
$$

In conclusion, the minimum value of the number of subsets $n$ is 15.