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?
+770 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
