Number of arrangements of 5 girls and 11 boys Q: Five girls and eleven boys are to be lined up in a row such that from left to right, the girls are in the order: $G_1, G_2, G_3, G_4, G_5$. In how many ways can this be done if $G_1$ and $G_2$ must be separated by at least 3 boys, and there is at most one boy between $G_4$ and $G_5$? A: 11!$\cdot$825
Apart from the solution from Five girls and eleven boys are to be lined up in a row such that... Where is my mistake? , is there an alternative approach via finding the number of integer solutions to the equation: $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 11$ for example? 
Here's my thought process. I viewed each girl $G_i$ as a "+" operator:
$$ G_1 \quad G_2 \quad G_3 \quad G_4 \quad G_5 $$
such that each gap between the $G_i$'s has a $x_i$, such that we have 
$$ x_1  + x_2 + x_3 + x_4 + x_5 + x_6 = 11$$
where 11 is the number of boys, and $x_i \geq 0$ for i $\in$ $\{1,3,4,5,6\}$, $x_2 \geq 3$, $x_5 \leq1$.
How do I go about this approach? I got stuck after reaching the last step stated above.
A: The constraint that $x_2\geq 3$ is easy to solve.  Just re-write your equation like: $x_1 + (3+x_2) + x_3 + x_4 + x_5 + x_6 = 11$ (now $x_2\geq 0$) and you can simplify.  Just think of starting off by forcing at least 3 boys into that position (you now have 8 left to work with).
The constraint that $x_4\leq 1$ is less easy.  Personally, I'd split this into cases: what happens when $x_4=0$? and what happens when $x_4=1$.  It is easy in this case because there are only two options.  It would be more painful if the constraint were something like $x_4\leq M$ where $M$ is a big number (presumably in the case where you had more than 11 boys).
