There is a tournament in table tennis (ping-pong) in which 8 players take place.

These are the rules of the tournament:

1) Each player plays with every other only 1 game.

2) If in the i-th round there is a game between A and B, and a game between C and D, and in the i+1-th round there is a game between A and C, then in the i+1-th there must be a game between B and D.

On how many different ways can we make a schedule for all rounds if it's not important which player plays on which table?

Question

Lea Gilmore

Mathematics

13 June, 13:37

There is a tournament in table tennis (ping-pong) in which 8 players take place.

These are the rules of the tournament:

1) Each player plays with every other only 1 game.

2) If in the i-th round there is a game between A and B, and a game between C and D, and in the i+1-th round there is a game between A and C, then in the i+1-th there must be a game between B and D.

On how many different ways can we make a schedule for all rounds if it's not important which player plays on which table?

+2

Answers (1)

Know the Answer?

Denise Norton · Answer 1

so 1 player playes 2345678 2 plays 345678 3 plays 45678 4 plays 5678 5 plays 678 7 plays 8 add those up and there is 26 total possible games and 13 rounds.