Is there a way to put all the 2-element subsets of $X = \{1, 2, 3, ... , n\}$ in 2 groups such that there is no 3-element subset $S \subset X$ for which all of its 2-element subsets are in the same group?
Pigeonhole Principle?
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combinatorics
discrete-mathematics
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0I'm doubtful that the pigeonhole principle sheds much light on this problem. In any case the current title won't help attract interested Readers as well as one more specific to your Question. – 2017-02-18
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0I think the key here is the size of $n$. Certainly if $n=3$ one does it easily. It seems impossible for sufficiently large $n$ by virtue of Ramsey's theorem. – 2017-02-18
1 Answers
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This is equivalent to looking for a complete graph where the edges (= two element subsets) cannot be coloured with two colours (=2 groups) such that there is no single-colour triangle (=3-element subset).
The corresponding Ramsay number says that for $n=6$ and above, there will always be a single-colour triangle, corresponding to a 3-element subset where all 2-elements subsets are in the same group.