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In my last test, our teacher gave us this question:

In an conference with 90 participants, the staff wants to divide all participants in 6 groups of 15 participants each. Each participant of each group must have at least one article with exactly 7 other participants of the same group. Prove using Graph Theory that this division is impossible.

I spent a lot of time thinking about, but couldn't find an definitive answer. I think that this involve bipartite matching somehow, but I have no idea on how to proceed with the proof. I thought that maybe creating a k-regular graph would prove that this don't have a matching, but I couldn't come with the graph representing the problem.

Can I get some help with this?

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  • $\begingroup$ The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! $\endgroup$
    – Raphael
    Commented Apr 17, 2018 at 20:47

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There is no 7-regular graph on 15 vertices, since $7\times 15$ is odd.

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  • $\begingroup$ Thank you! This is based on the handshaking lemma for graphs? $\endgroup$
    – LionsWrath
    Commented Apr 17, 2018 at 22:10
  • $\begingroup$ Exactly. That's the reason. $\endgroup$
    – Yuval Filmus
    Commented Apr 18, 2018 at 3:26

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