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Let $\mathcal F$ be a family of subsets of a set consisting of $n$ elements so that no element of $\mathcal F$ is a subset of another element. Prove that $\mathcal F$ can have at most $$\binom{n}{\Big\lfloor\frac{n}{2}\Big\rfloor}.$$

Are these partitions? I suspected I could use induction, but it isn't really clear.

Award task, Socialist Republic Croatia, Yugoslavia 1973

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  • $\begingroup$ Sure, I'll do that. $\endgroup$ – Orchid_2.718281828 Nov 10 '19 at 20:22
  • $\begingroup$ @ViktorGlombik I didn't want to write anything on my own because I translated it literally not to write anything wrong since I don't know if I have covered enough theory to conclude. $\endgroup$ – Orchid_2.718281828 Nov 10 '19 at 20:26
  • $\begingroup$ I mean, this is obvious, but I followed the author... $\endgroup$ – Orchid_2.718281828 Nov 10 '19 at 20:28
  • $\begingroup$ @ViktorGlombik I don't at all see how this would be equivalent to the sets in $\mathcal{F}$ being mutually disjoint. $\endgroup$ – Morgan Rodgers Nov 10 '19 at 20:45
  • $\begingroup$ Oh, now I see how much I still have to learn. $\endgroup$ – Orchid_2.718281828 Nov 10 '19 at 20:46
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This is just Sperner theorem.

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