# Thread: Probability: Placing people into groups

1. ## 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 = ${20\choose4,4,4,4,4} / 5!$

Also, is the division by 5! to negate the need of order? Because I originally thought that the ${20\choose4,4,4,4,4}$ would be the total result.

Anyways, I was thinking of starting the question something like this.
${19\choose4} * {16\choose4,4,4,4} /5!$

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

2. 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 = ${20\choose4,4,4,4,4} / 5!$
.
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 $A,B,C.$
Place them in three different teams . . . it doesn't matter which teams.

The teams look like this:
. . $\{A\:\_\:\_\:\_\}\quad \{B\:\_\:\_\:\_\}\quad \{C\:\_\:\_\:\_\}\quad \{\_\:\_\:\_\:\_\} \quad \{\_\:\_\:\_\:\_\}$

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

Hence, there are: . ${17\choose3,3,3,4,4} / 3!2!$ .ways the friends are separated.

Therefore, the desired number is: . ${20\choose4,4,4,4,4}/5! - {17\choose3,3,3,4,4}/3!2!$