How can I arrange a group of people at tables and switch them around so that no two ever meet twice?

Question

Say I have 5 tables of 4 people (i.e. 20 people in total) and a different game at every table. We play one round of games (so each table plays one unique game) and then switch tables. Is it possible to have every person play every game and no two people sit together for more than 1 round? I can show this is not possible for 25 people with 5 tables and 5 rounds but I'm not sure how the proof of whether it is possible works and how to actually work out the combinations.

NB: This is not a homework problem - I'm actually trying to do this as an event, but I'm also a mathematician so I'm curious about how it actually works! I can now find a solution where this is possible, which I found using trial and error, looking at patterns and solving a bit like a sudoku but I doubt this is a unique solution and I don't know what the formula is that would represent it.

Answer 1

It is possible! Here is one such schedule. Players are named with letters A to T, and games are numbered 1 to 5. I found this basically with brute force, taking symmetry into account to reduce the search space somewhat. In fact, up to permutation of players and rounds, I found that there were only six schedules fitting the requirements.

	A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S	T
Round 1	$1$	$1$	$1$	$1$	$2$	$2$	$2$	$2$	$3$	$3$	$3$	$3$	$4$	$4$	$4$	$4$	$5$	$5$	$5$	$5$
Round 2	$2$	$3$	$4$	$5$	$1$	$3$	$4$	$5$	$1$	$2$	$4$	$5$	$1$	$2$	$3$	$5$	$1$	$2$	$3$	$4$
Round 3	$3$	$4$	$5$	$2$	$3$	$5$	$1$	$4$	$4$	$5$	$2$	$1$	$5$	$1$	$2$	$3$	$2$	$4$	$1$	$3$
Round 4	$4$	$5$	$2$	$3$	$5$	$4$	$3$	$1$	$2$	$1$	$5$	$4$	$3$	$5$	$1$	$2$	$4$	$3$	$2$	$1$
Round 5	$5$	$2$	$3$	$4$	$4$	$1$	$5$	$3$	$5$	$4$	$1$	$2$	$2$	$3$	$5$	$1$	$3$	$1$	$4$	$2$