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Math Help - 2 Counting problems

  1. #1
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    2 Counting problems

    1.In how many ways can 12 persons seat themselves at 3 different round tables, 4 at each:

    (a) (12!) / (4!)^3
    (b) (12!)(3!^3) / (4!^3)
    (c) 12!
    (d) none of the above


    2. In how many ways can 12 persons be divided into 3 groups of 4 each:

    (a) (12!) / (4!)^3
    (b) (12!)(3!^3) / (4!^3)
    (c) 12!
    (d) none of the above
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  2. #2
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    Re: 2 Counting problems

    Quote Originally Posted by swordfish774 View Post
    1.In how many ways can 12 persons seat themselves at 3 different round tables, 4 at each:
    (a) (12!) / (4!)^3 (b) (12!)(3!^3) / (4!^3) (c) 12!
    (d) none of the above

    2. In how many ways can 12 persons be divided into 3 groups of 4 each:
    (a) (12!) / (4!)^3 (b) (12!)(3!^3) / (4!^3) (c) 12!
    (d) none of the above
    These are very nice problems.
    Do you plan to show any effort towards solving them?
    You know that this is not a homework service?
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  3. #3
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    Re: 2 Counting problems

    you are the second user to say that. (I posted another question, someone else said the same thing) I am here to just verify the answers and look for better methods

    1.
    selection for tables
    12C4*8C4*4C4 = (12!) / (4!)^3

    then you muliply 3! for circular permutations. so option b

    2. 12C4*8C4*4C4 = (12!) / (4!)^3

    then divide by 3! so option 4

    I came here to verify answers. and 2. question I am not sure because last option is 'none of these', so need to check if approach left some counts or if there was some double counting.
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  4. #4
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    Re: 2 Counting problems

    oo. Plato... you are the same user.
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