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How to solve this permutation question

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i49.tinypic.com/wmmhbl.png

    2. Relevant equations

    3. The attempt at a solution
    my answers :
    Q1:(a)
    (i) 8!
    (ii) 4!*5!
    Q1:(b)
    7P3*5P2
    Q2:
    4320 ways
    Q3:
    12!*3!
    Are my answers correct?
     
    Last edited: Dec 28, 2012
  2. jcsd
  3. Dec 27, 2012 #2
    Q1 looks fine. I haven't checked Q2, but Q3 is way off. How do you get that result anyway?
     
  4. Dec 27, 2012 #3

    haruspex

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    P or C?
    Doesn't seem enough. How did you arrive at that?
    The mugs of each colour are identical, and you only care about the pattern of colours here.
     
  5. Dec 28, 2012 #4
    Q2:
    6!*3!=4320 ways
     
  6. Dec 28, 2012 #5

    haruspex

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    That doesn't count the ways one set of mugs may be placed in relation to the other set.
     
  7. Dec 28, 2012 #6
    In Q.1 (b) you should use combinations and not permutations since we are talking about selections. for Q.2 try to think placing china mugs in some specific no. of allowed spaces and then rearranging them. In Q.3 you can't consider all mugs different. Mugs of same colour can't be rearranged within themselves.
     
  8. Dec 28, 2012 #7
    Ok so Q1(b) should be 7C3*5C2=350 ways right?
     
  9. Dec 28, 2012 #8

    haruspex

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    Yes, as I suggested in post #3.
     
  10. Dec 28, 2012 #9
    Q2:
    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 7P3 ways.

    Therefore, there are:61*7P3 possible arrangements.

    am i correct?

    and as for Q3 I don't know how to start
     
  11. Dec 28, 2012 #10
    Q2 looks fine to me now.

    Q3 contains a red herring, since first we're told that there are 3 identical red mugs, and then they should be regarded as one unit by being kept together. Imagine dumping the 3 red mugs in one large red box, then you have only 12 objects to arrange: the red box, 4 identical blue mugs and 7 identical yellow mugs.
     
  12. Dec 29, 2012 #11
    @Michael Redei
    Q3:
    1 red box + 4 identical blue mugs + 7 identical yellow mugs = 12!
    and there are 3 mugs in the red box so it should be 12!*3!
     
    Last edited: Dec 29, 2012
  13. Dec 29, 2012 #12

    Dick

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    Some of those mugs are identical! For example, there aren't 3! ways of arranging 3 identical red mugs. There's only one.
     
  14. Dec 29, 2012 #13

    haruspex

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    ... furthermore, the question does not ask for the number of ways of arranging the mugs. It asks for the number of ways of arranging the colours.
     
  15. Dec 29, 2012 #14
    Q3:
    12!/7!*4!=3960 correct?
     
  16. Dec 29, 2012 #15

    haruspex

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    Yes!
     
  17. Dec 29, 2012 #16
    It might look better as 12! / (7! 4!), or even better as $$\frac{12!}{7! \cdot 4! \cdot 1!} = 3960.$$
    12! / 7! * 4! would be (12! / 7!) * 4! = 2 280 960
     
  18. Dec 29, 2012 #17
    ok I got it and thanks for help everyone
     
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