If you have 48 people at a bar and they rotate…

I have a math question. I’m typing this out at an ungodly hour with my favourite homebrew of cocoa (with a secret blend of other ingredients, but mainly involving sugar) and I’m presented with an interesting math problem. I’m not the smartest mathematician out there and only know so much as my highschool syllabus (no, I’m not a university student yet). A combination of these factors makes me not able to answer this question. Here I am asking anybody interested if they could share their train of thought.

You have a room full of 48 people. You have 8 tables. They are divided into equal groups, thus 6 people sit at each table. They each have a discussion for the same length of time. After they finish, everybody on each table, save for one, will move to another table to have another discussion with a new group of people. This one person will stay at their initial table throughout the entire process. So in effect for each “round”, 5 people from each table will move to another. So in total there will be 8 rounds of discussions. They are not allowed to move to a table they have sat at previously. How do you determine the optimum movements such that the moving people mix with as many different people as possible?

… and of course, you must be able to prove that it is indeed the optimum movement. I did manage to come up with a solution, though I cannot say that it’s the optimum. My result was that every person would meet 28 new people (obviously they will meet at least 8 – those who stay at each table) in total throughout all the rounds – but perhaps somebody can tell me how to really tackle this problem?

Dion Moult

I've been developing software for well over 10 years, work as an architect (not the computer kind, the regular sort), and am classically trained as a pianist. I try to do the right thing when I get the chance in my field, such as through contributing to open-source communities and promoting sustainable living.

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