At least 2 girls between every pair of boys distribution question? Three boys and eight girls are seated randomly in a row of 11 chairs. All orders are equally probable. 
What is the probability that there are at least 2 girls between every pair of boys?
What is the simplest way that I can solve this problem? The part that stumps me the most is "at least 2 girls". 
 A: First off, clearly it doesn't matter if we assume the boys and girls are all different or all the same, so we assume all boys are identical, and all girls are identical.
We now create a pictorial representation for each arrangement: In this arrangement we place a bar for each boy and a star for each girl.
The following is a valid configuration:
$*|**|***|**$
How many valid configurations are there? If I tell you how many stars come before the first bar, how many stars come between the first and second bars, and after the fourth bar we are done, so we can express each valid configuration as a sequence of non-negative integers $(x,y,z,w)$, where $y$ and $z$ are at least two and the four numbers add $8$. Notice we can take each of these configurations, substract $2$ from $y$ and $z$. And we get a sequence of non-negative integers $(x',y',z',w')$ which add to $4$.
Conversely, given a sequence of non-negative integers $(x,y',z',w)$ which add to $4$, if you add $2$ to $y'$ and $z'$ you get a sequence of non-negative integers $(x,y,z,w)$, where $y$ and $z$ are at least two and the four numbers add $8$. So these two structures are in one-one correspondence. therefore the number of valid configurations is equal to the number of sequences of non-negative numbers $(x,y',z',w)$ that add to $4$, by stars and bars this is $\binom{7}{3}=35$
On the other hand there are $\binom{11}{3}=165$ configurations total.
The answer is hence $\frac{35}{165}=\frac{7}{33}\approx 21\%$
