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There are $6$ boys and $4$ girls in a class. How many ways are there to arrange them in a row if no girl stands next to each other?

I would know how to solve this if there are only $2$ girls. But since there are $4$ here, I'm stumped as how to proceed. The approach that I have tried is wrong, though I fail to see why. The answer I got from this approach is bigger than than $10!$

What I have tried is putting the girls in certain columns alternating with the boys, and I group a couple of girl and boy as $1$. When I did that, that left me with $6!$

I used $4! \cdot 6! \cdot 2 \cdot 6!$ And my answer is way off.

So, can someone help me point out the fault in my reasoning and help point me in the correct approach to tackle this question?

Thanks very much.

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1 Answer 1

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To position the girls among the boys so that no girl is adjacent to another, you must place at most one girl in the spaces that are between each boy, or at the ends of the line.

  • Count the ways to arrange the $6$ boys in a line (alone).

  • Count the ways to arrange the $4$ girls in a line (alone).

  • Count how to select $4$ of the $7$ spaces between, or at the ends, of the boys.


What I have tried is putting the girls in certain columns alternating with the boys, and I group a couple of girl and boy as 1. When I did that, that left me with 6!

That would be counting ways to arrange six distinct units.   But boy-1 does not have to always stand to the left of girl-1, so paired units don't have a fixed identity.

You'd need to count ways to arrange: (bg)(bg)(bg)(bg)(b)(b) where each round bracket is a unit, and the b and g are identical place holders for the boys and girls to stand on.   In any paired unit, the boy is always to the left of the girl to keep the girls separated.   To also allow for cases where a girl is at the start of the line, you need to add the count of: [g](bg)(bg)(bg)(b)(b)(b), where the square bracket must remain in its fixed position.

Adding these and multiplying by the ways to arrange boys and girls among their placeholders, the count is $$\left(\frac{6!}{4!2!}+\frac{6!}{3!3!}\right)\times 6!\times 4!$$

Which, after applying a combinatorial identity, will give the same answer you should obtain by my first method.

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  • $\begingroup$ Hey, thanks a lot for your help. I can understand your solution, though I still fail to see what's the problem with mine. To be more specific, why can't I multiply the 6! 4! With 2 and 6!? I multiplied with 6! because I regard the 6 as groups to be rearranged, and 2 because the girl and boy can switch sides. $\endgroup$
    – Lim LS
    Mar 10, 2015 at 2:16
  • $\begingroup$ @LimLS See the edit. $\endgroup$ Mar 10, 2015 at 2:50

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