Thread: Probability: Placing people into groups

Hello.
I am not too sure how to start this one.

Q: Three friends and 17 other boys are to be randomly divided into five
teams of 4 players. Out of the total amount of teams, how many will have at least 2 of our friends play for the same team?

ALL possible outcome total =

Also, is the division by 5! to negate the need of order? Because I originally thought that the would be the total result.

Anyways, I was thinking of starting the question something like this.


Does this sound about right? Thanks for any help, much appreciated.

Hello, chrisc!

Three friends and 17 other boys are to be randomly divided into five teams of 4 players.
Out of the total amount of teams, how many will have at least 2 of our friends
play for the same team?

ALL possible outcome total =
. Assuming the 5 teams are interchangeable, I agree!

I think I have an approach to this one . . .

The opposite of "at least two on the same team" is "none on the same team."
. . That is, the three friends are on separate teams.

In how many ways is this possible?


Call the friends
Place them in three different teams . . . it doesn't matter which teams.

The teams look like this:
. .


The other 17 players will be partitioned into: 3, 3, 3, 4, 4.

Hence, there are: . .ways the friends are separated.


Therefore, the desired number is: .