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Math Help - Problem on fundamental counting principle.

  1. #1
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    Problem on fundamental counting principle.

    There are 5 boys and three girls. If two girls cannot sit next to each other, how many ways can all of them sitin a row together.

    I'm not really sure if what I did was right, but my answer is 240. I added 5! (number if combinations the boys can make) and 6!/(6-3)! (the number of slots between the boys distributed to the number if girls) I got 120 for each. I'm not really sure if what I did was right, or even how to check if it's right.
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  2. #2
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    Re: Problem on fundamental counting principle.

    Quote Originally Posted by miguel11795 View Post
    There are 5 boys and three girls. If two girls cannot sit next to each other, how many ways can all of them sitin a row together.

    I'm not really sure if what I did was right, but my answer is 240. I added 5! (number if combinations the boys can make) and 6!/(6-3)! (the number of slots between the boys distributed to the number if girls) I got 120 for each. I'm not really sure if what I did was right, or even how to check if it's right.
    Your answer is really too low. There are two ways to approach this question.
    8!-2\cdot 7!=30240, that is subtracting the cases where they are together from the total.

    Or \frac{7!}{(7-2)!}\cdot 6!=30240. The six "others" make seven places to put the two girls.
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