[Math] Networking event rotation

combinatorics

In need of a little more math brain power. I am hosting an event where I ultimately want folks to meet new people every time we rotate and not be with the same person twice.

My problem is not having people stay at the same table while rotating. I can get 36 people in groups of 6 to rotate 3 times and never meet the same person, but some of the people never leave their original table or they come back to a table they have already been at. Each table has a different theme and game to play and I don't want people to play a game twice either.

I figure I need to either subtract the number of people or add more tables/games, but can't figure this out.

Best Answer

It is not hard to show that with $36$ people seated at six tables of six places each, there's no way to reseat everybody to a different table each time without pairing some of the same folks a second time. Consider the six people at the first table during the first seating. Where will they go for their second seating? If they must all go to different tables, one of them would have to remain at that first table (pigeonhole principle).

One idea I'm willing to propose involves $35$ people seated at seven tables of five places each. In this scheme we could have (up to) six rotated seatings, with no pair of people meeting more than once, and no person seated at the same table more than once. I note, however, that your preference was to have an even number ($4$ or $6$) of people at each table. What do you think about the proposed type of scheme?

28 People (Seven Tables of Four Each, Three Rounds without Repeats)

Given the desirability of an even number of people at tables (so that games which require a pair of teammates are conducted), I looked for a solution with seven tables and four people at each table (per round). Here's what I found:

First Round 

Table 1:  A1,B1,C1,D1
Table 2:  A2,B2,C2,D2
Table 3:  A3,B3,C3,D3
Table 4:  A4,B4,C4,D4
Table 5:  A5,B5,C5,D5
Table 6:  A6,B6,C6,D6
Table 7:  A7,B7,C7,D7

Second Round

Table 1:  A2,B3,C4,D5
Table 2:  A3,B4,C5,D6
Table 3:  A4,B5,C6,D7
Table 4:  A5,B6,C7,D1
Table 5:  A6,B7,C1,D2
Table 6:  A7,B1,C2,D3
Table 7:  A1,B2,C3,D4

Third Round

Table 1:  A3,B5,C7,D2
Table 2:  A4,B6,C1,D3
Table 3:  A5,B7,C2,D4
Table 4:  A6,B1,C3,D5
Table 5:  A7,B2,C4,D6
Table 6:  A1,B3,C5,D7
Table 7:  A2,B4,C6,D1

It is fairly easy to verify that all twenty-eight people are seated in each round, and that the three seatings at each table are free of repeats. Less easy to spot is whether there are any pairs of people who meet more than once. The design is based on a rotation of a certain kind, so that it actually suffices to check any one person (for duplicate meetings) to be sure that no one meets the same person twice. Here are the seatings for A1 (and you can see there are no repetitions):

Round 1:   A1,B1,C1,D1   (Table 1)
Round 2:   A1,B2,C3,D4   (Table 7)
Round 3:   A1,B3,C5,D7   (Table 6)

It follows that each participant will meet nine other people during this networking event.

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