Here's a fun math/Zoom question for you all: If I have 6 people in a group and each participant needs to meet each other in pairs of 2, how could I arrange Zoom Breakout rooms so that no one meets another twice? (3 Breakout rooms consisting of 2 people per session)
Here are the possible pairs I have (the letters represent initials of participants):
K x R, K x J, K x C, K x N, K x Z, R x J, R x C, R x N, R x Z, J x C, J x N, J x Z, C x N, C x Z, N x Z
So with this in mind, it should break down into 5 total sessions with 3 breakout rooms in each session and 2 participants per breakout room. I just can't seem to figure out how to configure it without duplicate people being in a session.
 A: What you're describing is the $1$-factorization (i.e., splitting into $1$-factors, which are perfect matchings) of a complete graph on $n$ vertices.
From the Wikipedia article on graph factorization:

One method for constructing a $1$-factorization of a complete graph on an even number of vertices involves placing all but one of the vertices on a circle, forming a regular polygon, with the remaining vertex at the centre of the circle. With this arrangement of vertices, one way of constructing a $1$-factor of the graph is to choose an edge $e$ from the centre to a single polygon vertex together with all possible edges that lie on lines perpendicular to $e$. The $1$-factors that can be constructed in this way form a $1$-factorization of the graph.

In other words, make a diagram like so, then consider its five rotations (rotate just the edges; keep the labels fixed):
$\hskip{3cm}$
A: 



Pair 1
Pair 2
Pair 3




AB
CD
EF


AC
BE
DF


AD
CE
BF


AE
CF
DB


AF
CB
DE




I solved it by approaching it as you would sudoko
