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Possibly wrong permutation/combination question?

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    In how many ways can 4 red balls, 3 white balls and 1 black ball be arranged in a line so that the black ball is always surrounded by a red and a white ball?

    Textbook answer: 20


    2. Relevant equations

    Chapter was on permutations and combinations, so
    ##n P r = n!/(n-r)!##
    ##n C r = n!/(r!(n-r)!)##

    3. The attempt at a solution

    We have 4 red, 3 white, 1 black.
    Black must be surrounded by 1 red and 1 white, so we treat RBW as one unit.
    There are two possible combinations of this unit, RBW and WBR, so the result will be multiplied by 2.
    There are 6 possible positions to place this 3 ball unit into an 8-ball space, so there are 6*2 = 12 different ways to place this unit.

    For the remaining, we regard only order of color, so we use nCr:
    For the red balls (only 3 remaining, and we have 5 spaces) there are 5 C 3 = 10 different ways to place them.
    For the white balls (only 2 remaining, and we have 2 spaces), there are 2 C 2 = 1 ways to place them.

    So the final result is : 12 * 10 * 1 = 120.

    I would like a confirmation that this answer is correct (I'm usually not, but maybe I am just this once :wink: )
     
  2. jcsd
  3. Apr 8, 2014 #2

    Dick

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    Sounds correct to me.
     
  4. Apr 13, 2014 #3

    Curious3141

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    I get 120 as well. This is ##\displaystyle \frac{6!}{3! 2! 1!} \times 2!## which is another way of looking at it.
     
    Last edited: Apr 13, 2014
  5. Apr 13, 2014 #4
    I also get 120.
    It is twice a permutation of 6 elements (as I'm grouping WBR and RBW) with 3 repetitions of one type and 2 of another type.
    [tex]2 \cdot P^{(3, 2)}_{6}=2 \cdot \frac{6!}{3!2!}=120[/tex]
     
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