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Math Help - How to solve this permutation question

  1. #1
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    How to solve this permutation question

    How to solve these two:
    http://i49.tinypic.com/72a8ll.png


    And is this question of combination?
    http://i50.tinypic.com/ehdb9j.png
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  2. #2
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    Re: How to solve this permutation question

    Hey bllnsr.

    It's a lot easier to show us what you have tried with these kinds of problems because if the intuition is wrong, then an attempt will show how it is wrong.

    Can you show us your attempts?
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  3. #3
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    Re: How to solve this permutation question

    Q2: it is written that china mugs are all separated from each other so I placed 6 plastic mugs and 3 china mugs alternately
    6!*3!=4320 ways

    Q3: This one confuses me and my answer to this question looks way off to me
    12!*3!

    and the last one I think is a question of combination :
    7C3*5C3=350 ways
    Last edited by bllnsr; December 28th 2012 at 01:02 AM.
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  4. #4
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    Re: How to solve this permutation question

    Hello, bllnsr!

    I can't see Q2, but I'll guess what the problem says.


    Q2: it is written that china mugs are all separated from each other
    so I placed 6 plastic mugs and 3 china mugs alternately: 6!*3!=4320 ways

    There are 6 different plastic mugs and 3 different china mugs to be placed in a row
    . . where no two china mugs are adjacent.

    Place the 6 plastic mugs in row, inserting spaces before, after and between them.
    . . . . \_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_

    There are 6! arrangements of the 6 plastic mugs.

    Choose 3 of the 7 spaces and place the 3 china mugs.
    . . . There are _7P_3 ways.

    Therefore, there are: . (6!)(_7P_3) \:=\:151,\!200 possible arrangements.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Suppose the 6 plastic mugs are identical and the 3 china mugs are, too.

    Place the 6 plastic mugs in a row with the spaces.
    Since they are identical, their order does not matter.

    Now select 3 of the 7 spaces and place the 3 china mugs.
    . . . There are _7C_3 ways.

    Therefore, there are: . _7C_3 \:=\:35 possible arrangements.
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