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Prove: Amongst any 6 people, there are 3 people pairwise acquainted or 3 people pairwise unaquainted.?
This is a question on combinatorial reasoning.
I have a vague idea of its solution.
But I want a systematic unassailable reasoning
which can be explained to a beginner in
combinatorics.
@gooseknows : 2nd part of your last sentence
says "...3 friends are mutually enemies...."
Do you really want to use the word ' friends '
here ?
1 Answer
- 1 decade agoFavorite Answer
i'll say 'friends' and 'enemies' instead of 'aquainted' and 'unaquainted'
pick any person, call him A. since there are 5 other people and every two people are either friends or enemies, he will have at least 3 friends or 3 enemies (you can justify this by pigeonhole principle, but it is clear that it is true). suppose without loss of generality that he has 3 friends. if any two of the three are friends with each other, then we have the required group of 3 friends by taking the two and A. otherwise, the 3 friends of A are mutually enemies (pairwise, amongst themselves) and we have the required group of 3 enemies
might be helpful to draw a graph...
