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Math Help - probability

  1. #1
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    probability

    How many different ways can a family of 6 sit in a row at the movies if the 2 youngest children must sit next to each other?
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  2. #2
    MHF Contributor arbolis's Avatar
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    First, there are 5 ways the 2 children can be sit. In each case the rest of the family can be sit in 4! different ways. So I think the answer is 5.4!. And it's equal to 120. (Which is 5! So now I see that we could have think the problem as if the 2 children were one person of the family and not 2).
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  3. #3
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    Hello, Help09!

    I got a different answer . . .


    How many different ways can a family of 6 sit in a row at the movies
    if the 2 youngest children must sit next to each other?
    Duct-tape the two youngest together.
    . . Then we have five "people" to arrange: . \boxed{AB}\;C\;D\;E\;F
    And they can be arranged in: . 5! = 120 ways.

    But in each of those arrangements, \boxed{AB} can be replaced by \boxed{BA}

    Therefore, there are: . 2 \times 120 \:=\:{\color{blue}240} ways.

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  4. #4
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    Let these be the seats

    --- --- --- --- --- ---

    We first fix the first two seats for the youngest children

    [--- ---] --- --- --- ---

    There are 4! ways of filling the other 4 seats

    Similarly for the cases,

    --- [--- ---] --- --- ---

    ... so on for 5 cases, each of which contributes a 4!. So the total number of ways is 5*4!=120 * 2 = 240 (where 2 children can be swapped)

    In general for n seats and k people in connesecutive seats,
    [(n-k+1)(n-k)!]k! = (n-k+1)!k!

    Same as our answer: 5!.2!=240
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  5. #5
    MHF Contributor arbolis's Avatar
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    I'm sorry for my answer, Soroban and Sarim are right. Multiply my result by 2.
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