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    liang11's Avatar
    liang11 Posts: 4, Reputation: 1
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    #1

    Jul 24, 2009, 11:27 PM
    Permutations and combinations
    The numbers 1,2,3,4,5,6 are arranged in a circle. How many ways can this be done if
    1) at least 2 odd numbers are together
    2) the numbers 5 and 6 are together but 4 is not next to 5 or 6
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Jul 25, 2009, 01:09 AM

    1. The number of possible arrangements without restrictions are 6! / 6 = 120.

    Now, consider a group of odd numbers. You have only 1, 3 and 5. You can put them into groups 1, 3; 3, 5 or 1, 5, each giving 2 possible arrangements.

    Now, take a group of odd numbers (1, 3), and treat them like a separate item. You therefore will have 5 items total, namely (1, 3), 2, 4, 5, 6.

    The number of possible arrangements of the items is found to be 5! / 5 = 24.

    Total possible arrangements = 24 x 2 x 3 = 144.

    2 because you have 2 different combinations for the group of odd numbers, and 3 because you have 3 different groups containing different odd numbers.

    2. Now, 5 and 6 are together: arrangements = 2
    Total items = 5 (again)

    (5, 6), 1, 2, 3, 4

    Now, 4 must not be put next to (5, 6) you can therefore have:

    1, (5, 6), 2, 3, 4
    1, (5, 6), 3, 2, 4
    2, (5, 6), 1, 3, 4
    2, (5, 6), 3, 1, 4
    3, (5, 6), 1, 2, 4
    3, (5, 6), 2, 1, 4

    1, 2, (5, 6), 3, 4
    1, 3, (5, 6), 2, 4
    2, 1, (5, 6), 3, 4
    2, 3, (5, 6), 1, 4
    3, 1, (5, 6), 2, 4
    3, 2, (5, 6), 1, 4

    That makes 12 other possible arrangements (don't forget that they are arranged in a circle.)

    So, number of combinations = 2 x 12 = 24

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