Counting problem

There are 9 ways to choose a single person.

You then have \(8 \choose 4\) ways to choose a group of 4 people. But, you have to divide by 2, to correct for overcounting.

Once that is done, you have \({4 \choose 4} = 1\) way to choose the remaining. 

Thus, there are \(9 \times {8 \choose 2} \div 2 = \color{brown}\boxed{315}\) ways.

All credit goes to Melody: https://web2.0calc.com/questions/a-group-of-nine-people-is-be-split-into-three-teams_1

We can choose  any 4 of the 9   people to be on  1 of the 3 teams

And then we can choose any 4 of the 5 remaining people to be on 1 of the remaining 2 teams

The last person is on the last team by default

So

C ( 9,4)  * C ( 3,1) * C(5,4) * C (2,1)  =   3780  ways

Not totally confident about this, but BuiderBoy  will  probably have the correct answer  !!!!  (LOL !!! )