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The answer to the first question is no"no." Suppose $|M|=11$ and $\mathcal S\subset\binom M5$ and $|\mathcal S|=77.$ Since each $5$-set contains five $4$-sets, and since $77\cdot5=385\gt330=\binom{11}4,$ by the pigeonhole principle some $4$-set must be contained in two members of $\mathcal S,$ which are both contained in the same $6$-set.

The answer to the first question is no. Suppose $|M|=11$ and $\mathcal S\subset\binom M5$ and $|\mathcal S|=77.$ Since each $5$-set contains five $4$-sets, and since $77\cdot5=385\gt330=\binom{11}4,$ by the pigeonhole principle some $4$-set must be contained in two members of $\mathcal S,$ which are both contained in the same $6$-set.

The answer to the first question is "no." Suppose $|M|=11$ and $\mathcal S\subset\binom M5$ and $|\mathcal S|=77.$ Since each $5$-set contains five $4$-sets, and since $77\cdot5=385\gt330=\binom{11}4,$ by the pigeonhole principle some $4$-set must be contained in two members of $\mathcal S,$ which are both contained in the same $6$-set.

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bof
bof
  • 13.4k
  • 2
  • 43
  • 66

The answer to the first question is no. Suppose $|M|=11$ and $\mathcal S\subset\binom M5$ and $|\mathcal S|=77.$ Since each $5$-set contains five $4$-sets, and since $77\cdot5=385\gt330=\binom{11}4,$ by the pigeonhole principle some $4$-set must be contained in two members of $\mathcal S,$ which are both contained in the same $6$-set.