Guests at a Round Table

Fifteen chairs are evenly placed around a circular table. On the table are the name cards of fifteen guests. After the guests sit down, it turns out that none of them is sitting in front of his own card. Prove that the table can be rotated so that at least 2 guests are simultaneously correctly sitted.

Solution


|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Fifteen chairs are evenly placed around a circular table. On the table are the name cards of fifteen guests. After the guests sit down, it turns out that none of them is sitting in front of his own card. Prove that the table can be rotated so that at least 2 guests are simultaneously correctly sitted.

The table has 15 possible positions. One of this positions, viz., the initial one, has no man matching a card, leaving 14 positions for possible matchings. There are 15 people and 14 table positions. By the pigeonhole principle, there bound to be a position for which at least to men match their names.

Clearly, the result holds for any number of guests.

References

  1. A Decade of the Berkeley Mathematical Circle, The American Experience, Volume I, Z. Stankova, Tom Rike (eds), AMS/MSRI, 2008, pp. 165-168

Related material
Read more...

  • Six integers out of 10: Pigeonhole Principle
  • Students in Classes
  • The Math Lotto
  • Boys and Girls Compete in Maths
  • Pigeonhole in Paint Distribution
  • Increasing or decreasing subsequence
  • Guests at a Round Table
  • Membership in a Math Circle

  • |Contact| |Front page| |Contents| |Up|

    Copyright © 1996-2018 Alexander Bogomolny

    71546441