Answer to Question #112857 in Combinatorics | Number Theory for Shelley Guo

Question

In a baseball league of 8 teams, how many games are needed to complete the schedule if each team plays 8 games with each other?

Answer 1

Imagine that there are only two teams. In this case, they will play only one game.

When you add a third team, the number of games will increase by two - a new team will play with each of the two existing ones.

When you add a fourth team, the number of games will increase by three - a new team will play with each of the three existing ones.

Etc.

When adding the m-th team, the number of games will increase by (m-1) - a new team will play with each of the (m-1) existing ones.

If teams play only one game with each other then the total number of games between m teams:

$1 + 2 + ... + (m-1) = \displaystyle \sum_{n=1}^{m-1}n$

If teams play only one game with each other then the total number of games between 8 teams:

$1 + 2 + ... + 7 = \displaystyle \sum_{n=1}^{7}n=\frac{1+7}{2}\cdot 7=28.$

In case of 8 games for each pair the total number of games will be

8 * 28 = 224