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,

There must be at least two black balls after each of the first $m - 1$ white balls in the row, so the answer is zero unless $n - m \geq 2(m - 1) \implies n \geq 3m - 2$.  
Method 1:  Set aside $2(m - 1) = 2m - 2$ black balls.  That leaves us with $m$ white balls and $n - m - (2m - 2) = n - 3m + 2$ black balls to arrange.  Choose $m$ of the $m + n - 3m + 2 = n - 2m + 2$ positions for the white balls, which can be done in 
$$\binom{n - 2m + 2}{m}$$
ways.  Finally, insert two black balls to the immediate right of the first $m - 1$ white balls to ensure that there are at least two black balls between any two white balls.   
I thought of the simpler method shown above after posting my first attempt at the problem.  My original approach is shown below.
Method 2:  We can represent each arrangement of black and white balls by a bit string in which a white ball is represented by a $1$ and a black ball is represented by a $0$.  We consider cases depending on the position of the $m$th white ball.
Case 1:  The $m$th white ball is in the last position.  
We have a bit string of length $n$ that ends in $1$.  We must fill the first $n - 1$ positions of the bit string with the remaining $m - 1$ ones and $n - m$ zeros.  Since each of the first $m - 1$ white balls must be followed by at least two black balls, we must arrange $m - 1$ blocks of the form $100$ and $n - m - 2(m - 1) = n - 3m + 2$ zeros.  The number of arrangements of this form is equal to the number of ways we can choose which $m - 1$ of the $m - 1 + n - 3m + 2 = n - 2m + 1$ objects will be filled with blocks of the form $100$, which is 
$$\binom{n - 2m + 1}{m - 1}$$ 
Case 2:  The $m$th white ball is in the next to last position.
Since no two white balls can be consecutive, the last position must be filled with a black ball.  Hence, we have a bit string of length $n$ that ends in $10$.  We must fill the first $n - 2$ positions of the bit string with the remaining $m - 1$ ones and $n - m - 1$ zeros.  Since the first $m - 1$ white balls must be followed by two black balls, we must arrange $m - 1$ blocks of the form $100$ and $n - m - 1 - 2(m - 1) = n - 3m + 1$ zeros.  The number of bit strings of this form is equal to the number of ways we can choose which $m - 1$ of the $m - 1 + n - 3m + 1 = n - 2m$ objects will be filled with blocks of the form $100$, which is 
$$\binom{n - 2m}{m - 1}$$
Case 3: The $m$th white ball is not in the last two positions.
In this case, every white ball is followed by at least two black balls.  Thus, we must arrange $m$ strings of the form $100$ and $n - m - 2m = n - 3m$ zeros.  The number of arrangements of this form is equal to the number of ways we can choose which $m$ of the $m + n - 3m = n - 2m$ objects will be filled with blocks of the form $100$, which is 
$$\binom{n - 2m}{m}$$
Total:  Since the three cases are mutually exclusive and exhaustive, we can find the total number of arrangements by adding the number of arrangements of the three cases above.
\begin{align*}
\binom{n - 2m + 1}{m - 1} & + \binom{n - 2m}{m - 1} + \binom{n - 2m}{m}\\
& = \binom{n - 2m + 1}{m - 1} + \binom{n - 2m + 1}{m}\\ 
& = \binom{n - 2m + 2}{m}
\end{align*}
where we have applied Pascal's Identity twice.
