Maths solution for students in pairs puzzle

First off, I have to warn, I’m not much of a maths wizz, so if I’ve messed up any terms, I apologise (except for ‘maths’, which is the correct Australian way of saying ‘math’ ;))

I was once chatting with someone who was doing a Phd in Mathematics, and he mentioned that he’d solved a problem that a friend of his (who was a teacher) was having, using the branch of mathematics he was expert in (which I think was something like set theory).

The problem is as follows:

There are twenty students in a class. Students are to be paired up for a week, and then rotated, so the following week everyone is working with a different partner. No two students are to be paired together more than once for the nineteen weeks the class is together.

For example, if there were four students:
week 1:
student 1 and 2 work together; student 3 and 4 work together

week 2:
student 1 and 3 work together; student 2 and 4 work together

week 3:
student 1 and 4 work together; student 2 and 3 work together

I can work it out for ten students by hand, but there’s supposedly a mathematical way of doing it, which can be extended to very large numbers of students. Do any of the wise folk here have any suggestions?

Never apologise for being correct - Math ith a church thervith for Catholicth!! :wink:

I must admit that I don’t really understand your question - are you asking for a way of easily producing a listing of all the different combinations of pupils that will occur during the 19 week course?


You’re essentially talking about “round robin scheduling”, which is commonly applied to sporting events in which each team (or participant) plays every other team (or participant) exactly once.

This page discusses some of the math in detail, and you can set N=20 to get a full schedule of pairings of 20 students.

Perhaps I don’t understand the problem correctly.

I take it as rather simple. If N=number of players, then (N-1)! are the number of pairings.

If each plays once a week, then the number of weeks would be
N-1 if the number of players was even and simply N if odd.

Bryan, thanks for that link - it’s exactly what I was after. Now I just have to wade through the maths :slight_smile: