You have 10 players all set to play a round robin tournament where everyone will play everyone else once. The standard 10-player Berger table will be used.

10 players:
Rd 1: 1-10, 2-9, 3-8, 4-7, 5-6.
Rd 2: 10-6, 7-5, 8-4, 9-3, 1-2.
Rd 3: 2-10, 3-1, 4-9, 5-8, 6-7.
Rd 4: 10-7, 8-6, 9-5, 1-4, 2-3.
Rd 5: 3-10, 4-2, 5-1, 6-9, 7-8.
Rd 6: 10-8, 9-7, 1-6, 2-5, 3-4.
Rd 7: 4-10, 5-3, 6-2, 7-1, 8-9.
Rd 8: 10-9, 1-8, 2-7, 3-6, 4-5.
Rd 9: 5-10, 6-4, 7-3, 8-2, 9-1.

Here’s the challenge: There are 10 seats available for the players to sit at, boards 1 through 5 and each board has white and black. Is there a way to have each player not sit in the same seat more than once?

Failing that, is there a way to have each player not sit at any of the tables more than twice, irrespective of color?

Each time I get to one of these 10-player round robins I always try to rotate the seating to ensure everyone maximum fairness throughout the event but I invariably get a player sitting at a table 3 times and so if there are any math geniuses out there who know a solution, I’d be happy to hear it.

Thank you!