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Thread: Probability

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    Junior Member 1729's Avatar
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    Probability

    How do you do part (iii) for this question? Thanks in advance

    Screen Shot 2018-06-23 at 1.19.17 pm.png

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    Member kawaiipotato's Avatar
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    Re: Probability

    Count the ways to fix the boys in a circle first. Then count the ways to place the girls between the boys.
    This will be your desired outcomes, and your total outcomes is just 7!.
    Roy G Biv and 1729 like this.
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    Junior Member HeroWise's Avatar
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    Re: Probability

    But the question states "Chosen"

    So if thats the case and we choose from the original 6 boys and 6 girls shouldnt you get this?

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    Re: Probability

    Quote Originally Posted by HeroWise View Post
    But the question states "Chosen"

    So if thats the case and we choose from the original 6 boys and 6 girls shouldnt you get this?
    The question's wording is such that it wants us to assume we have already picked 4 boys and 4 girls.
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    Re: Probability

    I see, i thought it was a carry over

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    Re: Probability

    i) 12C8
    ii) 6C4 x 6C4
    iii) 4!3! / 7!

    Draw a circle with 8 spots.

    We 'fix' in 1 boy. 3 boys left. Now, we're going to place the boys making sure to leave a seat empty between each boy. 3 boys can go in 1st of those seats, 2 in the second, 1 in the last.

    Now we will fill in the blanks with the girls. 4 girls can go in first of the empty spots, 3 in the second, etc.
    therefore: 4! x 3!

    Since they've asked us the probability, we need to answer in terms of total possible ways to arrange these 8 students in a circle, which is 7!.

    Hence 4!3! / 7!

    Note: HeroWise has done this visually in his/her drawing above
    Last edited by Roy G Biv; 24 Jun 2018 at 12:02 PM.

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