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Question on permutation and combination(girls and boys standing in a line)

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data


    http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w03_qp_6.pdf
    question 6 b ii
    2. Relevant equations



    3. The attempt at a solution
    The concept is bothering me. the answer is 5! × 3! × 4C1. Why is it a permutation problem? Plus why do we use 5! × 3! in a permutation problem, isn't it used in combinations? as 5!X 3! shows the nos of ways it can be arranged 'in any order' So how can we use that in a permu where order DOES matter!

    The reason I think its a permutation problem is because of http://answers.yahoo.com/question/index?qid=20110530100750AAXrtRp
    other than that I dont relly know why it is a permutation problem even.
     
  2. jcsd
  3. Mar 23, 2012 #2

    Dick

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    Draw a picture. How many ways can you position a block of 5 boys in a line of 8 people with 3 girls? Then given that number of patterns how many ways can you shuffle the boys and girls around in their positions?
     
  4. Mar 23, 2012 #3
    I don't think you got what I was trying to say. Yes i know 'how' it is 5! X 3! ways, but i don't know 'why' it is a permutation, because they are clearly being selected randomly...
     
  5. Mar 23, 2012 #4

    Dick

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    Ok, so you know the solution. If it's any help, I would not have written 4C1 instead of 4. The answer is just 4*5!*3!. Why you would write 4C1 instead of 4 escapes me. Is that what you are asking?
     
  6. Mar 23, 2012 #5
    No, ok firstly, do you agree its a permutation sum and NOT a combination?
     
  7. Mar 23, 2012 #6

    Dick

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    I agree that it has somewhat more to do with permutations, but I also don't think there is an absolutely clear separation between permutation problems and combination problems. Sometimes you have to use a little of both.
     
  8. Mar 23, 2012 #7
    I hesitate to do this since I'm not well-versed in combinations/permutations, but don't we have more than 4(5!)(3!)? The 5! comes from the block of boys, and the 3! from a block of girls, but aren't there more different combinations of girls around the boys? 1 girl in front, 2 in back, 2 in front 1 in back, 3 in front 0 in back. I'm wondering myself how those are accounted for (not to hijack the thread...).
     
  9. Mar 23, 2012 #8
  10. Mar 24, 2012 #9
    You forget another one, 3 in the back 0 in front.
    <--
    GBGG or 3!5!
    GGBG or 3!5!
    GGGB or 3!5!
    BGGG or 3!5!
    =4.3!5!

    Anymore combination you can add?
     
    Last edited: Mar 24, 2012
  11. Mar 25, 2012 #10
    Ah, got it. Thanks.
     
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