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Math Help - Seating arrangements

  1. #1
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    Seating arrangements

    The question has four parts:
    A. Suppose there are 8 seats in a row with 8 people to seat in the seats, one per seat, how many seating arrangements are possible?

    B. Suppose 2 of the 8 are twins and are indistinguishable, thus any arrangement should be labeled the same regardless of which twin in is which seat, how many arrangements are possible?

    C. Same as B but with triplets instead of twins

    D. Same as B but with triplets and a different set of twins, still 8 total.

    For A it should be 8! right? idk what to do for B, C, and D though, any help would be appreciated, TIA.
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  2. #2
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    Quote Originally Posted by jason0 View Post
    Suppose there are 8 seats in a row with 8 people to seat in the seats, one per seat, how many seating arrangements are possible?
    C. Same as B but with triplets instead of twins
    If we rearrange the string "AAABCDEF" is that not the exact same as part C?
    The number of possible arrangements is \frac{8!}{3!}.
    Now you show us the rest.
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  3. #3
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    Quote Originally Posted by Plato View Post
    If we rearrange the string "AAABCDEF" is that not the exact same as part C?
    The number of possible arrangements is \frac{8!}{3!}.
    Now you show us the rest.
    so would it be like
    part a is ABCDEFGH so 8! = 40320 combinations
    part b is AABCDEFG so 8!/2! = 20160 combinations
    part c is AAABCDEF so 8!/3! = 6720 combinations
    part d is AAABBCDE so 8!/(3!+2!) = 3360 combinations?
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  4. #4
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    Quote Originally Posted by jason0 View Post
    part d is AAABBCDE so 8!/(3!+2!) = 3360 combinations?
    That should be \frac{8!}{(2!)(3!)}
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  5. #5
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    Quote Originally Posted by Plato View Post
    That should be \frac{8!}{(2!)(3!)}
    thank you
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