Three boys and nine girls are seated randomly in a row of $12$ chairs.

What is the probability that there are at least $2$ girls between every pair of boys?

My attempt: To find the numerator, we count how many ways we can have at least $2$ girls between every pair of boys. Observe the following diagram:


The dashes represent the boys and the dots represent the girls that must be between them at minimum.

We have $3!$ ways to order the boys, and $\binom 944!$ ways to choose 4 girls to satisfy the restriction and order them among the dots. Now we have $9-4=5$ girls that we can distribute on the far left, inner left, inner right, or far right. The number of ways we can do this is the number of nonnegative integer solutions to $$x_1 + x_2 + x_3 + x_4 = 5$$ which is $\binom{5+4-1}{4-1}=\binom 83$.

Therefore we have $3! \cdot \binom 944!\cdot \binom 83$ as our numerator, and our probability is $$\frac{3! \cdot \binom 944!\cdot \binom 83}{12!}$$

It appears that this is incorrect. Where did I go wrong?

  • $\begingroup$ Girls are not identical $\endgroup$ – NewGuy Feb 22 '18 at 18:46
  • $\begingroup$ @NewGuy That means I ought to multiply by $5!$ then, right? $\endgroup$ – Tiwa Aina Feb 22 '18 at 19:09

You need to create arrangements of boys ('$b$'s) and girls ('$g$'s) first. At the moment we are still treating them as identical.

You must, therefore, begin with the following scenario if there have to be at least two girls between every pair of adjacent boys:


Then the remaining 5 '$g$'s can go in the 4 gaps between 3 '$b$'s in $\binom{5+3}{3}$ ways, this is equivalent to counting non-negative integer solution vectors to $x_1+x_2+x_3+x_4=5$.

All we do now is multiply by the ways the 9 girls and 3 boys can replace their letters '$g$' and '$b$'. The are 9!3! ways to do that. Since there are 12 children we have probability:


I'm sorry to say that you made a couple of mistakes but the main one was choosing and ordering the 4 girls first to go in between the boys. By doing that you opened yourself up to all sorts of overcounting.

Trying to produce a correct solution once you had made that mistake would have been pretty tough, so don't feel too bad about it.


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