I can’t seem to figure this problem out. I really would appreciate it if someone would help me. A set of football matches is to be organized in a “round robin” fashion, meaning every participating team plays a match against every other team once and only once. If 91 matches are played, how many teams participated?
Honors Algebra 2 Bonus-Help
- Thread starter Trenchest
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Hello, and welcome to FMH! 
If \(n\) is the number of teams and \(M\) is the number of matches played, then these two quantities are related by the formula:
[MATH]M=\frac{n(n-1)}{2}[/MATH]
This comes from the fact that we are looking at the number of ways to choose 2 from \(n\) objects:
[MATH]{n \choose 2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}[/MATH]
Can you proceed?
If \(n\) is the number of teams and \(M\) is the number of matches played, then these two quantities are related by the formula:
[MATH]M=\frac{n(n-1)}{2}[/MATH]
This comes from the fact that we are looking at the number of ways to choose 2 from \(n\) objects:
[MATH]{n \choose 2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}[/MATH]
Can you proceed?
HallsofIvy
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Another way to look at it is that every one of the "n" teams has to play every one of the "n-1" games. That would be n(n- 1) games but since each game counts for two teams, it is only \(\displaystyle \frac{n(n-1)}{2}\).