A tennis coach divides her 9-player squad into three 3-player groups with each player in only one group. How many different sets of three groups can be made?
For the first team, the coach has \(9\) possible choices and picks \(3\) players. This is the same as \(\dbinom{9}{3}\) as order doesn't matter in a group. Thus, there are \(\dbinom{9}{3} = 84\) ways to pick the first team.
For the second team, the coach has \(6\) possible choices left and again picks \(3\) players. This is \(\dbinom{6}{3}\), due to the reasoning above. Thus, there are \(\dbinom{6}{3} = 20\) ways to pick the second team.
For the third team, the coach only has \(3\) choices for players, and there is only way to choose 3 people from 3 people. Thus, there is \(\dbinom{3}{3} = 1\) to pick the third and final team.
Multiplying it all together, we get \(84 \cdot 20 \cdot 1 = \fbox{1680}\) ways! :D