Ok here we go, Use the following double counting to stablish some lemmas. Call $m_k$ partial bound for the max number of subsets such that we cant find six $A_i$ with the desired property. First, count the number of sets of $6$ elements such that $5$ of them belong to one $A_i$, using this for $n=7$, we get that for $n=7$, $5{\binom{7}{6}}> 2m$, so we get $m_7=17$. Now counting the number of $7$ elements set, such that $5$ of them belong to an $A_i$, for $n=8$, we get $m_8=45$, and Finally for the other $n$, counting the number of $8$ element sets, $5$ of them in an $A_i$, we get a better bound that the desired one.

I really dont understand the solution Could someone explain more please.

Who can write a clearly solution?

Good people, pls !What is the official solution for this interesting problem?

Anyone?....

I don't believe the first solution works and here's a rewrite of what they meant. Define a six perfect set as such a set of $A_{i_1}, \dots, A_{i_6}$. Claim: Let $m_k$ be the maximal value of $m$ for $n = k \ge 6$ that such no six perfect set is present and let $n > k$. Then \[ \binom{n-5}{k-5} \cdot m_n \le m_k \cdot \binom{n}{k} \]Proof: Consider the bipartite graph from the $5$-tuples $A_1, \dots, A_m$ to the set of $k$-tuples $B_1, B_2, \dots$ where we connect $A_i$ to $B_j$ if $A_i \subset B_j$. Then for any $k$-tuple $B_j$, its degree must be at most $m_k$ else a six perfect set occurs in it. Then we just multiply by the degree of each $5$-tuple. This condition simplifies as saying that for $n \ge k$ \[ \frac{m_n}{n(n-1)(n-2)(n-3)(n-4) } \le \frac{m_k}{k(k-1)(k-2)(k-3)(k-4)}. \]Then we always want to consider $k = n-1$ as to maximize the strength of our bound because $m_k$ is integral, which makes this $m_n \le \frac{m_{n-1} \cdot n}{n-5}.$ Since $m_6 = 5$, we get that $m_7 = 17, m_8 = 45$.

Does exist a combinatorial solution,without the graphs theory?

Please....Is possible any solution without the graphs theory?Thx in advance

Bump.....

This is problem from《A Path to combinatorics for undergraduates》by Titu Andreescu, Zuming Feng. See exercise $7. 13$.

Thanks.....