Counting chair arrangements with multiple gaps This question asked how many ways there are to seat $m$ people in a line of $n$ chairs so that no two sit next to each other. I was wondering about a generalization: how many ways are there to place $m$ people in $n$ chairs so that each of them are separated by at least $k$ empty spaces?
I don't really know where to start. Some approaches involving recursion come to mind, but they quickly become overcomplicated.
 A: We count the number of choices of chairs. If we are then concerned with who sits where, we multiply by $m!$.
Line up  $m$ special chairs in a row. These are the chairs the people will actually sit on.
Put $k$ ordinary chairs in each of the $m-1$ gaps between people. Of course we need $n\ge k(m-1)$.
That leaves $n-k(m-1)-m$ ordinary chairs. They are to  be distributed in $m+1$ places, the $m-1$ gaps between special chairs, and the two ends to the left and right  of all the special chairs.
Counting the number of ways to do this is standard Stars and  Bars:
$$\binom{n-(m-1)k-m+m+1-1}{m+1-1}.$$
A: Note: Added answer due to lack of exposition in the one by Danny C.

For $i \in [1:m+1],$ let $x_i$ be the gap between the $i^\text{th}$ and $(i-1)^\text{th}$ occupied chair: $x_1$ carries the interpretation of being the number of chairs left empty in the beginning of the line, and $x_{m+1}$ the number left empty at the end. These gaps have to sum up to the total number of chairs minus $m$, the number of chairs occupied.
$x_1 + x_2 + \dots + x_m + x_{m+1} = n - m$,
such that for $i \in [2:m] , x_i \ge k$. 
For $i \in [2:m]$, let $x_i = k + y_i$, $x_1 = y_1, x_{m+1} = y_{m+1}$. Then $y_i \ge 0$, and the above equation changes to:
$y_1 + y_2 + \dots + y_m + y_{m+1} = n - m - k(m-1)$.
The number of solutions to this is $\binom{n  - k(m-1)}{m}$ by the stars and bars method.
Lastly, permute the $m$ people in the $m$ selected chairs to get the answer as:
$\binom{n  - k(m-1)}{m}$$m!$
A: Very interesting question, after giving it some thought I was thinking about this: let the chairs be numbered $1$ to $n$ and let $a_1,a_2,\dots,a_m$ be the (numbered) chairs that has people sitting on there. 
Thus, $1\leq a_1<a_2-k<a_3-2k<\cdots<a_m-(m-1)k\leq n-(m-1)k$, where every integer $a_i,a_j$ satisfies $|a_i-a_j|>k$.
Therefore the equation is simply $n-(m-1)k\choose m$.
(if anyone notice any errors please point out ..)
