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Arrangements in circles

  1. Jun 8, 2005 #1
    How many orders can we have of 3 red roses and 3 white roses arranged in a circle? Apparently the answer is 10 - but how do yiou work it out? I know that in a circle of "n" different objects the number of different arrangements is (n-1)!, but what about the repeats?

    If anyone could explain a proven method I would be most greatful.

    Thanks in advance.. :rofl:
     
  2. jcsd
  3. Jun 9, 2005 #2
    Well let me explain the logic behind the (n-1)! In order to count how many orders can be created, you have to fix one of the roses, and calculate how many orders can be created using the (n-1) roses. The answer is (n-1)! of course. To understand the problem's solution, the following would help (WR is a White Rose, and RR is a Red Rose),

    WR / RR-WR-RR-RR-WR. This is just one of the orders that can be created. It doesn't matter which rose you fix. At first glance you can say that 5!=120 orders can be created. But there is a slight thing that you miss. All of the same colored roses are identical, which makes you count some orders twice, or more times. In order to avoid this you divide (n-1)! by 3!(the number of orders created using the roses of the same color) and 2!(the number of orders created using the remaining 2 roses - mind that we fixed a rose.). So in short 5!/3!2!= 10.

    Hope this helps...

    Cheers,
    Can
     
  4. Jun 9, 2005 #3

    mathman

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    I am puzzled by the way the problem is proposed - specifically, arranged in a circle. If two arrangements are considered the same if you can rotate one to match the other, I can visualize only 4 different possibilities.

    rrrwww
    rrwrww
    rrwwrw
    rwrwrw

    ???
     
  5. Jun 9, 2005 #4
    wisredz, I do not see why you would have to fix one.
    The number of different orders if they were align in a row would be:
    [tex]\frac{n_{total}!}{n_{red}!*n_{white}!}[/tex]

    But if they are arranged in a circle the number of orders would be:
    [tex]\frac{n_{total}!}{n_{red}!*n_{white}!*(n_{total}-1)}[/tex]

    6!/(3!*3!*5) = 4
     
  6. Jun 10, 2005 #5
    well, In order to count how many orders you can create, you have to have a fixed point. That is why we have the formula (n-1)!. But you are right :) In what I did, I count some positions more than once.
    Anyway, I still have problem understanding why you divide by 5 to find how many orders can be created around a circle. Could you please explain it?
     
  7. Jun 10, 2005 #6
    From each circular arrangement of 6 roses you can make 6 different linear arrangements because you can choose each of the six roses as a start for the linear arrangement.
    ...
    ehh....
    and then I thought, but I want 1 of those 6 so there are 5 that I do not want so I divide by 5... but I now see that this reasoning is wrong, even though it accidentally gave the right answer
     
  8. Jun 11, 2005 #7
    Yes the reasoning is wrong, but the funny point is that it always gives the right answer :)
     
  9. Jun 11, 2005 #8
    n_total = 2
    n_red = 1
    n_white = 1

    n_total!/(n_red!*n_white!*(n_total-1))
    = 2

    Shouldnt that be 1?

    -- AI
     
  10. Jun 11, 2005 #9
    Yup, I didn't try for 2. 4 gives the right answer. Any good idea how to do it?
     
  11. Jun 11, 2005 #10
    As far as i recall(Something i read back some time back (probably it was Hall & Knight)), there is no closed form solution for objects which are arranged in circular fashion and the objects are repeated. However, if there is atleast one object in that set which isnt repeated then we can **fix** that object, and take permutations of the remaining.

    -- AI
     
  12. Jun 11, 2005 #11

    mathman

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    To illustrate the difficulty with this problem, I worked out explicitly what happens for two sets of two and two sets of three.

    For two sets of two, the linear permutations are:
    rrww, rwwr, wwrr, wrrw (group 1)
    rwrw, wrwr (group 2)
    When considering a ring, the group 1's collapse together and the group 2's collapse together. Note that there is no obvious way to get from 6 linear to 2 ring permutations.

    For two sets of three the ring permutations are:
    rrrwww (1)
    rrwwrw (2)
    rrwrww (3)
    rwrwrw (4)
    To get the linear permutations, (1),(2),(3) can start at any of the six positions to get a different result. However (4) has only one other alternative - start with w. This gives the total of 20 possible linear permutations.

    Conclusion - no easy formula!
     
  13. Jun 13, 2005 #12
    Can u please explain that?

    -- AI
     
  14. Jun 13, 2005 #13

    mathman

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    As you can see from the examples, some ring permutations can generate different linear perturbations from any place on the ring, while others (like the last one in each example above) do not have this property.
     
  15. Jun 13, 2005 #14
    Ohhhhhh!! NOW i get it!
    You meant to say each of the ring permutations need not produce the same number of linear permutations! D'oh should've noticed that! Thanks for explaining.

    -- AI
     
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