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Math Help - Permutations/Combinations

  1. #1
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    Permutations/Combinations

    Determine the number of ways that the 12 members of the boys' basketball team can be lined up if Joe, Taner and Josh must all be together?

    My answer:

    10! x 2

    The answer they have is 604800, which is derived from 10!/3!
    How is this a permutations with identical terms? No where in this question does it say that Joe, Taner, and Josh are triplets. What's wrong with my thinking?
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  2. #2
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    I'm getting a different answer.

    Treat the 3 people who must be together as 1 person. Then there are 10! ways to arrange them. Within this group of 3, there are 3! ways for them to be arranged. so I get 10! x 3!.
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  3. #3
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    EDIT:

    I meant that I used 10! x 3! also, but they have 10!/3!
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  4. #4
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    Hello, skeske1234!

    I don't agree with their answer.


    Determine the number of ways that the 12 members of the boys' basketball team
    can be lined up if A, B\text{ and }C must all be together?

    My answer: 10! x 2

    The answer they have is 604,800, which is derived from 10!/3! ??

    Duct-tape A,B,C together.
    . . Note that there are 3! = 6 ways they could be ordered.

    Then we have 10 "people" to arrange: . \boxed{ABC}\;D\;E\;F\;G\;H\;I\;J\;K\;L
    . . and there are 10! ways.


    Therefore, there are: . 10! \times 6 possible line-ups.

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  5. #5
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    Oh man I was sitting here for like 10 minutes trying to figure out just how the hell they managed to get that. I had the exact same thought, "Why the heck would you divide by the permutation of the boys and not multiply!"
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