What is the number of ways of selecting four persons in a row of $12$ so that no two persons sitting next to each other are selected? 
$12$ persons are arranged in a row.  What is the number of ways of selecting four persons so that no two persons sitting next to each other are selected?

I tried solving this but I am finding too many cases for this problem. Is there a shorter way to solve this problem ?
 A: We will arrange eight blue balls and four green balls.
Line up eight blue balls.  This creates nine spaces, seven between successive blue balls and two at the ends of the row.
$$\square \color{blue}{\bullet} \square \color{blue}{\bullet}\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$
In order to separate the green balls, we choose four of these nine spaces in which to place a green ball.  For instance, if we choose the first, third, sixth, and eighth spaces we obtain
$$\large{\color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet}}$$ 
Now number the balls from left to right.  The numbers on the green balls represent the locations of the four selected people.
A: Hint 1. The problem is equivalent to count the number of non-negative integer solutions of the equation
$$x_1+(x_2+1)+(x_3+1)+(x_4+1)+x_5=12-4$$
that is
$$x_1+x_2+x_3+x_4+x_5=5$$
($x_1$ is the number of persons on the left of the first selected person, $(x_2+1)$ is the number of persons between the first and the second selected persons...).
See Stars-and-bars technique.
Hint 2. Select, without any restriction, $4$ persons among the first $9$ persons from the left, say $1\leq n_1<n_2<n_3<n_4\leq 9$. Then choose the persons at the positions $n_1$, $(n_2+1)$,  $(n_3+2)$, and $(n_4+3)$. 
