Number of binary strings with $n$ ones and $m$ zeros $f(n,m)$ is the number of binary strings with up to $n$ ones and up to $m$ zeros.
Prove that the number of possible strings is: $${n+m+2 \choose n+1} -1$$
I got to the point that:
$$\sum_{a=0}^n \sum_{b=0}^m {a+b \choose a}$$
And I also understand that there are $(n+1)$ options for the amount of ones and $(m+1)$ options for the amount of zeros.

We can find the sum by generating functions
\begin{align*}
  \sum_{b=0}^{m} \binom{a+b}{b} &= [x^m] \frac{1}{(1-x)}\frac{1}{(1-x)^{a+1}} \\
  \implies \sum_{a=0}^{n} \sum_{b=0}^{m} \binom{a+b}{b} &= [x^m]\sum_{a=0}^{n}\frac{1}{(1-x)^{a+2}} \\
  &= [x^m]\left(\frac{1}{x\left(1-x\right)^{n+2}}-\frac{1}{x\left(1-x\right)}\right) \\
    &= \binom{m+n+2}{m+1}-1
\end{align*}
