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Math Help - Permutations &Combinations in Probability

  1. #1
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    Permutations &Combinations in Probability

    8 people of different heights are to be seated in a row. what is the probability that (a) there are at least 3 people sitting between the tallest and shortest? (b) the tallest and shortest are sitting next to each other?


    Hi, if you could please explain how to do the problem above I would really appreciate it! Stats is my weak point in math and I would really appreciate if you could explain the logic behind the work you did, if not its okay! Thanks anyway!
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  2. #2
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    Total ways to seat these 8 people: 8!

    a) Let's count how many ways we can seat our tallest and shortest with at least 3 people between them. I'm going to have the tallest to the left and shortest to the right. We'll double our count because I could have an equal number of representations in reverse.

    3 space configs: TxxxSxxx, xTxxxSxx, xxTxxxSx, xxxTxxxS
    Each config has 6! ways to seat the remaining people; 4*6! ways total.

    4 space configs: TxxxxSxx, xTxxxxSx, xxTxxxxS
    Each config has 6! ways to seat the remaining people; 3*6! ways total.

    5 space configs: TxxxxxSx, xTxxxxxS
    Each config has 6! ways to seat the remaining people; 2*6! ways total.

    6 space configs: TxxxxxxS
    This config has 6! ways to seat the remaining people; 6! ways total.

    It appears we have 10*6! ways to do this, remember to double. So we've got 20*6! ways to handle this.

    Probability: (20*6!)/(8!) = 20/56 = 5/14


    b) I think this one is much easier to count. Use a similar method to what I've done above and good luck!
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