Speed Networking Seating Schedule

Why 6 tables with 9 seats each? This allows for 54 attendees, and you only have 45.

It's not possible with more seats per table than tables for everyone to see everyone else only once.

Take the nine people at table 1 for example: during the second round, spread everyone from table 1 to other tables. You have three people leftover with no seat away from the people they saw the first round. For that to work, I'd do 5 seats at 9 tables.

Originally Posted by jsaragosa

I also need an answer to this question. I'm looking for a random team generator - 4 teams x 10 weeks. Each person must be in a team with each other person at least once.

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Is this really solvable ?

8 people are around a table, the second round they don't want to hear each other again, so each needs to go to a different table, but there are only 6 tables?

round 1
person 1 Table one
person 2 Table one
person 3 Table one
person 4 Table one
person 5 Table one
person 6 Table one
person 7 Table one
person 8 Table one


round 2
person 1 Table one with 7 others
person 2 table two with 7 others
person 3 table three with 7 others
person 4 table four with 7 others
person 5 table five with 7 others
person 6 table six with 7 others
person 7 ?
person 8 ?


How can they not hear someone again where do persons 7 and 8 go, or am I misunderstanding something? The key is that attendees only see each other 1 time at most.

to solve, tables needs to be considerably more than people at a table, otherwise if it is only once you hear a commercial you will never get a solution. tables =people-1 easy, you just rotate everyone every minute

and it is a very old post if i read the dates!