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In a party there are $10$ people. Total $14$ handshakes were made . Prove that there exists two different groups of some people(more than 3 each and one may be common in both) in which each person in the group is having handshake from two persons from that group.

My idea was this for a general N vertices $N \geq5$ :

  • Consider a graph of $N$ vertices and $N+4$ edges . If we can show that there exist two cycles sharing no edges then we are done most likely , but how do we show it ? I am not getting the idea for that
ProblemDestroyer
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  • I don't think the original statement to be proven is actually true. You could have 5 people forming a complete graph (with 10 edges) and another 5 people forming a line graph or tree graph (with 4 edges). I don't see how these 10 people could contain two such groups. Or are the groups allowed to overlap? Are the two cycles allowed to share edges, but just not be identical? – Jaap Scherphuis Aug 29 '22 at 09:33
  • @JaapScherphuis your graph does have two cycles of 3 sharing no edges in the completed one . Oh yes also i forgot mentioning there can be one person in common in both groups . In general my main problem is to solve the graph frame problem , i think thats always true – ProblemDestroyer Aug 29 '22 at 09:48

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