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Kürschák Math Competition
1960 Kurschak Competition
1
1
Part of
1960 Kurschak Competition
Problems
(1)
Among any 4 people at a party
Source: 1960 Hungary - Kürschák Competition p1
10/10/2022
Among any four people at a party there is one who has met the three others before the party. Show that among any four people at the party there must be one who has met everyone at the party before the party
combinatorics