Consider the set $S$ (whose elements are indexed by $i\in\{1,...,\binom{l}{m}\}$) of bit strings of length $l$ that contain exactly $m$ $1$s and $l-m$ $0$s. For each string $s_i\in S$, define $x_{i,j}$ to be the number of $0$s between the $j^{th}$ $1$ and $(j+1)^{th}$ $1$ in $s_i$. Now consider the distribution of all the $x$s ($x\in\{0,...,l-m\}$). I have a formula (for which I have ample evidence empirically) stating the frequency of each possible value of $x$: $(l-x-m+1)\binom{l-x-1}{m-2}=(m-1)\binom{l-x-1}{m-1}$.
Why is the count of the object I've described equal to $(m-1)\binom{l-x-1}{m-1}$?
Here's an example where $l=5$ and $m=3$. $S=\{00111,01011,01101,01110,10011,10101,10110,11001,11010,11100\}$. $s_1=00111$ contributes two $0$s to $x$'s frequency count (since $x_{1,1}=0$ and $x_{1,2}=0$), $s_2=01011$ contributes a $1$ and a $0$ to the frequency count (since $x_{2,1}=1$ and $x_{2,2}=0$), while $s_5=10011$ contributes a $2$ and a $0$ to the frequency count (since $x_{5,1}=2$ and $x_{5,2}=0$). My formula states that there are $12$ instances of $x=0$, $6$ instances of $x=1$, and $2$ instances of $x=2$, which is all correct.