Combinatorics - binary - very interesting question I tried to solve this problem without any success.
I really hope that you will know the answer.
$f(n,m)$ presents the number of binary strings (empty string included) that include at most n times '1', and at most m times '0'.    
Prove:
$$f(n,m) = \binom{n+m+2}{n+1}-1$$
 A: Here's the answer. Consider the number $g(n,k)$ of binary strings with exactly $k$ zeros (and at most $n$ ones). These strings can have $0,1,\ldots,n$ ones, i.e., they may look as follows
$$\underbrace{0\cdots 0}_{k \text{ times}},\qquad \underbrace{0\cdots 0}_{k \text{ times}}1,\qquad \underbrace{0\cdots 0}_{k \text{ times}}11,\:\ldots,\:\underbrace{0\cdots 0}_{k \text{ times}}\underbrace{11\ldots 1}_{n \text{ times}}$$
Of course, each time, we may distribute the $k$ zeros among a total of $k+j$ digits, for $j=0,\ldots,n$. Thus, this number is given as
$$g(n,k)=\sum_{j=0}^n\binom{k+j}{k}$$
It is well-known that (**)
$$\binom{k}{k}+\binom{k+1}{k}+\cdots+\binom{k+n}{k}=\binom{k+n+1}{k+1},$$
and thus, 
$$g(n,k)=\sum_{j=0}^n\binom{k+j}{k}=\binom{k+n+1}{k+1}=\binom{k+n+1}{n}.$$
Now, your strings may have between $0$ and $m$ zeros. Thus, the solution to your problem is
$$f(n,m)=\sum_{k=0}^m g(n,k)=\sum_{k=0}^m\binom{k+n+1}{n}=\binom{n+1}{n}+\cdots+\binom{n+m+1}{n}$$
and applying (**) again yields
$$f(n,m)=\binom{n+m+2}{n+1}-\binom{n}{n}=\binom{n+m+2}{n+1}-1.$$
EDIT To prove (**), you can telescope. Note that $\binom{k+j}{k}=\binom{k+j+1}{k+1}-\binom{k+j}{k+1}$, so
$$\sum_{j=0}^n\binom{k+j}{k}=\sum_{j=0}^n(\binom{k+j+1}{k+1}-\binom{k+j}{k+1})=\binom{k+n+1}{k+1}-\binom{k}{k+1}=\binom{k+n+1}{k+1}.$$
