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Math Help - Probability: Placing people into groups

  1. #1
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    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.
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  2. #2
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    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!

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