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Math Help - Circular Permutations

  1. #1
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    Circular Permutations

    Hi guys, need with with this question:

    In how many ways can 5 gentlemen, involving Roy and Ben, and 5 ladies, including Jane, be seated at a round table, such that - Jane sits next to neither Roy nor Ben?

    This is how I did it:
    Case 1 - Jane sits next to Roy
    Grouping Jane and Roy, no.of ways = (9!/9) * 2! = 80640

    Case 2 - Jane sits next to Ben
    No.of ways = (9!/9) * 2! = 80640

    Taking complement,
    (10!/10) - 2(80640) = 201600

    Is my method wrong? Thanks in advance!
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  2. #2
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    Re: Circular Permutations

    Quote Originally Posted by Blizzardy View Post
    In how many ways can 5 gentlemen, involving Roy and Ben, and 5 ladies, including Jane, be seated at a round table, such that - Jane sits next to neither Roy nor Ben?

    This is how I did it:
    Case 1 - Jane sits next to Roy
    Grouping Jane and Roy, no.of ways = (9!/9) * 2! = 80640

    Case 2 - Jane sits next to Ben
    No.of ways = (9!/9) * 2! = 80640

    Taking complement,
    (10!/10) - 2(80640) = 201600

    Is my method wrong? Thanks in advance!
    Your method is correct but your counting is off.
    Recall that \|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|, where \|A\| is the number of cases in A.
    You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.
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  3. #3
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    Re: Circular Permutations

    Quote Originally Posted by Plato View Post
    Your method is correct but your counting is off.
    Recall that \|A\cup B\|=\|A\|+\| B\|-\|A\cap B\|, where \|A\| is the number of cases in A.
    You found the first two but not the case in which Jane is seated between Ben & Roy. i.e. BJR or RJB.
    Thanks! But to calculate case 3, do I just group BJR together?
    i.e. No. of ways (Case 3) = (8!/8) * 2! = 10080 (Only arrange B and R)
    So, taking complement, (10!/10) - 2(80640) - 10080 = 191520
    Is this correct?
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  4. #4
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    Re: Circular Permutations

    Quote Originally Posted by Blizzardy View Post
    Thanks! But to calculate case 3, do I just group BJR together? i.e. No. of ways (Case 3) = (8!/8) * 2! = 10080 (Only arrange B and R) So, taking complement, (10!/10) - 2(80640) - 10080 = 191520
    I have not done the actual calculations. But the method is correct.
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