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A teacher has 6 girls and 8 boys to arrange in for a choir. Determine the number of ways she can arrange the 14 children in a single row if no two girls should stand next to each other.

How do you do this? I just need to know how to solve this. You can use another example to explain this to me because this question is very tricky to me .

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  • $\begingroup$ Try 2 girls and 1 boy. Then 2 girls and 2 boys etc. to determine what the constraints are, and how many arrangements are possible satisfying the constraints. $\endgroup$ – Math Lover Sep 28 '17 at 22:25
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Strategy:

  1. Arrange the eight boys in a row.
  2. This creates nine spaces, seven between successive boys and two at the ends of the row in which we can place the girls.
    $$\square b \square b \square b \square b \square b \square b \square b \square b \square$$ To separate the girls, choose six of these spaces in which to place the six girls.
  3. Arrange the six girls in the chosen spaces.
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    $\begingroup$ Am i correct to say that: Place all 8 boys in a row. There are 8! ways to do this. There are now 9 positions into which the 6 girls can go: | B | B | B | B | B | B | B | B | There are P(9,6) ways of placing the 6 girls into these 9 positions. 8! * 9!/3! = 2,438,553,600 $\endgroup$ – Mapoz Sep 28 '17 at 23:25
  • $\begingroup$ @Mapoz That is correct. $\endgroup$ – N. F. Taussig Sep 28 '17 at 23:26

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