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Ways to arrange rose bushes

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Mr. Flowers plants ten rosebushes in a row. Eight of the bushes are white and two are red, and he
    plants them in random order. What is the probability that he will consecutively plant seven or more
    white bushes?


    3. The attempt at a solution

    correct set up for the answer is 9/ 10 choose 2

    There are 9 ways to plant the rose bushes with 7 or more white in a row.
    7w w r r
    7w r w r
    7w r r w
    r r 7w w
    w r 7w r
    r w 7w r
    r r w 7w
    r w r 7w
    w r r 7w

    At first, I thought the answer will be 9/10!, since there are 9 different possible combinations where there are at least 7 whites in a row, and of the 10 rose bushes, there are 10! ways to shift move them around.

    I know the answer is to divide 10 choose 2 red bushes by 9, but I dont know why.
    [
     
  2. jcsd
  3. Apr 8, 2013 #2

    CompuChip

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    You are nearly correct, except that you didn't account for the fact that in each of your 9 possibilities, you can interchange the whites and reds. For example, if I number the white bushes 1 to 8 and the red bushes a and b, then both
    1 2 3 4 5 6 7 8 a b
    and
    3 5 8 2 1 4 6 7 b a
    would fall under the first "7w w r r" possibility, but they are counted as distinct possibilities in the total of 10!. So if you multiply your 9 by the number of permutations of the white bushes, and the number of permutations of the red bushes, you will get the correct number of possibilities.

    Once you get the correct answer, if you play around with the factorials a bit, you can get the (10 choose 2) to appear.
     
  4. Apr 9, 2013 #3

    haruspex

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    semidevil, did you understand Compuchip's answer? In case not, think of it that way...
    You correctly listed the 9 possible arrangements, using a scheme in which all whites are considered indistinguishable and all reds likewise. But the 10! would be if considering each bush unique. Instead of that, you need to continue to treat bushes of the same colour as identical, i.e. the count of all possible sequences of 8 ws and 2 rs. That's a matter of choosing which two of the 10 are rs: 10 choose 2.
     
  5. Apr 9, 2013 #4
    Just to make sure I understand. It won't be 10!, because 10! Is the number of ways I can shift all the bushes, so each white is distinct. I understand and feel comfortable of why 10! Is not correct.

    10 choose 2 means I am choosing 2 of the red roses right? Why is not not 10 choose 7 or 10 choose 8?

    Still a bit confused there.
     
  6. Apr 9, 2013 #5

    CompuChip

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    10 choose 2 is the same as 10 choose 8: choosing 2 items is the same as picking 8 items you DON'T choose.
    If you know how to write 10 choose 2 in factorials you can see this: in 10 choose 2 = 10! / (2! 8!) you can interchange the 2 and 8 to get 10 choose 8 = 10! / (8! 2!) which is of course the same because the order of the multiplication is irrelevant .

    Also note that in that expression you see the 10! pop up again, but it is divided by the number of ways you can shuffle the set of 8 and through set of 2 identical objects, i.e. 8! and 2!, respectively.
     
    Last edited: Apr 9, 2013
  7. Apr 9, 2013 #6

    haruspex

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    No, it's the number of ways of choosing which of the 10 positions will be allocated to the two red roses.
     
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