Arrange black and white balls so that each pair of white balls is separated by at least two black balls I am trying to solve the following question: 
How many linear arrangements of $m$ white balls and $(n-m)$ black balls are possible such that each pair of white balls is separated by at least two black balls? The white balls are indistinguishable among themselves and the black balls are indistinguishable among themselves.
Please provide the solution to this.
Thanks,

Let $x_0,x_1,\dots,x_m$ represent the number of black balls between the respective white balls.  Specifically, $x_0$ is the number of black balls to the left of the first white ball.  $x_1$ is the number of black balls between the first and second white ball, etc...
$\underbrace{\bullet}_{x_0}\circ\underbrace{\bullet\bullet\bullet}_{x_1}\circ\underbrace{\bullet\bullet\bullet}_{x_2}\circ\underbrace{\bullet\bullet\bullet\bullet}_{x_3}\circ\dots\circ\underbrace{\bullet\bullet\bullet}_{x_m}$
We know there are a total of $n-m$ black balls.  Thus, $x_0+x_1+\dots+x_m=n-m$
Further, the condition that there must be at least two black balls between every white ball implies that for $i\in\{1,2,\dots,m-1\}$ you must have $x_i\geq 2$.
We recognize first that there is a bijection between valid arrangements of balls and solutions to the system.  We try then to count how many integral solutions there are to the system:
$\begin{cases}x_0+x_1+\dots+x_m=n-m\\ 0\leq x_0\\ 2\leq x_1\\ 2\leq x_2\\ \vdots\\ 2\leq x_{m-1}\\ 0\leq x_m\end{cases}$
Make a change of variable:  $y_0=x_0, y_m=x_m, y_i=x_i-2$ for all other $i\in\{1,2,\dots,m-1\}$
This is then the system:
$\begin{cases}y_0+y_1+\dots+y_m = n-3m+2\\ 0\leq y_0\\ 0\leq y_1\\ \vdots\\ 0\leq y_m\end{cases}$
This is of a known problem form.  See for example this recent related post.  Note that in this scenario it is as though there are $m+1$ buckets and $n-3m+2$ balls yet to place.
