Each person can be considered as a set, with at most 3 elements (wines) Prove by contradiction:www.ddhw.com Assume no three people drink the same wine, i.e., for any three out of the total nine sets, their intersection is empty. First, we can prove that there exist two disjoint sets. Consider any four sets A,B,C,D,E, given the assumption that no three sets have non empty intersection, A need to have at least 4 elements in order to have non-empty intersection with each of B,C,D,E.www.ddhw.com Now, suppose A n B = empty, put together, A u B has at most 6 elements. Now, added in another set C, then either A n C or B n C is non empty, i.e. (A u B) n C is non empty. Now, consider yet another set D, (A u B) n D is non empty, but by assumption, ((A u B) n C) n ((Au B) n D) is empty, in anothe word, each added set 'will take away some different wines' from A u B, and this kind of robbery can not go on beyond 6 times, but there are 7 robbers around, impossible. Can some one give an example that the conclusion will not hold for 8 people? www.ddhw.com
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