0

We figured out how to play 15 games with 5 players but not six yet. 3 teams of 2, changing players each game. One team sits out each game. We need the team sequenced so it works out that everyone is teamed at least once with each of the other 5 players. Here is what 5 players looks like with the single person sitting out. Thanks in advance. Al

1  Ken  Terry     -     Dean  Deron            Al
2  Terry  Dean   -     Al  Deron                 Ken
3  Al   Ken         -     Dean  Deron           Terry
4  Terry  Ken     -     Al  Deron                  Dean
5  Terry  Dean   -     Al  Ken                     Deron
6  Terry  Dean   -     Deron  Ken               Al
7  Al   Terry        -     Dean  Deron            Ken
8  Ken  Dean     -     Al  Deron                  Terry
9  Terry  Deron  -     Al  Ken                      Dean
10 Al  Dean       -     Ken  Terry                 Deron
11 Ken  Dean    -     Terry  Deron              Al
12 Deron  Terry -     Al  Dean                    Ken
13 Al  Dean       -     Ken  Deron               Terry
14 Al  Terry        -     Ken  Deron               Dean
15 Ken  Dean    -     Al  Terry                     Deron
Arthur
  • 199,419
Alan
  • 1
  • 1
  • 1
    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 18 '23 at 21:50
  • Using tag "combinatorial designs" means that you have already encountered this subdomain. Therefore, why do you consider this problem "from scratch" ? Don't you know results/algorithms ? – Jean Marie Nov 18 '23 at 23:03
  • I added the combinatorial-designs tag. Probably should have commented, sorry. – Karl Nov 18 '23 at 23:28
  • 2 players playing 2 other players with 2 sitting out. Changing teams each game so no two sets of players play the same team again. Just like the example given for 5 players but adding one more to make six players. With 5 players one person sits out every 5th game. Hope this helps. – Alan Nov 20 '23 at 00:52

0 Answers0