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Math Help - Probability

  1. #1
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    Probability

    There are 9 cars and 9 parking spaces in a row. If 3 sport cars are randomly in the group of 9 cars, what is the probability that the sports cars are parked in a row?

    For this question is found the number of ways the sports cars can be parked in a row in 9 spots. They can be parked in 7*6 different ways out of a possible 9!.. correct?

    So i had 42/362880=.00011574
    Doesnt this seem a little low?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathlete2 View Post
    There are 9 cars and 9 parking spaces in a row. If 3 sport cars are randomly in the group of 9 cars, what is the probability that the sports cars are parked in a row?

    For this question is found the number of ways the sports cars can be parked in a row in 9 spots. They can be parked in 7*6 different ways out of a possible 9!.. correct?

    So i had 42/362880=.00011574
    Doesnt this seem a little low?
    for each of the 42 ways of parking the 3 sports cars there are 6! arrangements of the remaining cars, so the required probability is:

    (42 6!)/9! ~= 0.0833

    RonL
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  3. #3
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    Hello, mathlete2!

    Another approach . . .


    There are 9 cars and 9 parking spaces in a row.
    If 3 sport cars are randomly in the group of 9 cars,
    what is the probability that the sports cars are adjacent?

    There are: . 9! possible arrangments.


    Duct-tape the 3 sports cars together.
    . . Then there are 7 "cars" to arrange: . \boxed{SSS}, X, X, X, X, X, X
    There are: . 7! permutations.

    But the 3 sports cars can be permuted in 3! ways.

    Hence, there are: . (7!)(3!) ways for the sports cars to be adjacent.


    Therefore, the probability is: . \frac{(7!)(3!)}{9!} \;=\;\frac{1}{12}

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