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The Conversing Club (Posted on 2003-12-29) Difficulty: 4 of 5
There are 10 tables in the Conversing Club and 15 members. Each day, 3 people sit together around each of 5 of 10 possible tables in the club talking to each other.

Every week (7 days) everyone is at the same table with everyone else exactly once. Also, nobody is at the same table twice in the course of a week to provide a change of scenery each time. The first day is as following:


(The second day A couldn't sit with B, or C; B couldn't sit with C; D couldn't sit with E or F, but could sit with A, B, or C.)

How could their schedule be configured?

(Based on Fifteen Schoolgirls)

See The Solution Submitted by Gamer    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Solution (no computer program used) | Comment 2 of 6 |
(In reply to Solution (no computer program used) by Penny)

What was the manner of solution? Are there other solutions? Can it be done with fewer than 10 tables? (Yes, those are separate tables, not separate rooms.)
  Posted by Charlie on 2003-12-30 10:22:35

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