Number of series of combined zeroes and ones Hello I've this exercise that I am struggling with and would love to know how to approach it. 
Give the number of $m$ element series composed of $n$ ones(1) and $m-n$ zeros(0), where no one(1) neighbors the other.
For example:
$m = 3$, $n = 2$, only correct solution is $101$
$m = 3$, $n = 1$, correct are $100$, $010$, $001$
I notice that there must be $n$ or $n-1$ series of zero fillers because ones can't neighbor. But how can I count number of arrangements of these zero fillers? How to approach this?
The correct answer for this exercise is probably ${m-n+1 \choose n}$. Could you explain why exactly?
 A: This is a classic stars and bars problem.  Put the $m-n$ zeros in a row (the stars).  Now you have $m-n+1$ slots for ones, including the ends of the row.  You have to choose $n$ of them to actually put the ones in.
A: Let's break this down into two easier problems:  


*

*Find the number of $m$ element series composed of $n$ ones and $m-n$ zeros where no ones neighbor each other and the last element is zero.

*Find the number of $m$ element series composed of $n$ ones and $m-n$ zeros where no ones neighbor each other is one, and the last element is one, so the next-to-last element is zero.
To solve the first: Each one must have a zero to its right, and if you pair the one and the zero as a unit, then that zero "protects" against any one having a zero to its left. So we have $n$ "units" and $m-2n$ left-over zeros, with $m-n$ total items (units or zeros). This is
$$ \binom{m-n}{n}$$
To solve the second: Wipe out the last one, and you are left with the same problem as before, only instead of $n$ ones you have $n-1$ ones, and instead of $m$ total numbers you have $m-1$.This is
$$ \binom{(m-1)-(n-1)}{n-1} = \binom{m-n}{n-1}$$
Finally, add these together. We use the "addition formula"
$$\binom{r}{k} = \binom{r-1}{k} + \binom{r-1}{k-1}$$
with $k=n$ and $r-1=m-n$:
$$\binom{m-n}{n}+ \binom{m-n}{n-1}  = \binom{m-n+1}{n}$$
A: Maybe you can find a recursive formula.
Let $F_m$ be the number of possible such sequences of length $m$, and let $x=(x_1,\ldots,x_m)$ be such a sequence.
If $x_1=0$, then you can fill out the remaining part $(x_2,\ldots,x_m)$ of the sequence in $F_{m-1}$ ways, because when $x_1=0$, there are no restrictions on the subsequent elements.
If $x_1=1$, then you must have that $x_2=0$, because you would otherwise violate the defining property of the sequence. But then you can choose the remaining elements $(x_3,\ldots,x_m)$ in $F_{m-2}$ ways.
In total, $F_m = F_{m-1}+F_{m-2}$.
