Combinatorics-number of permutations of $m$ A's and at most $n$ B's Prove that the number of permutations of $m$ A's and at most $n$ B's equals $\dbinom{m+n+1}{m+1}$.
I'm not sure how to even start this problem.

Here's a combinatorial solution. Take a string $\alpha$ containing exactly $m+1$ A's and $n$ B's. Identify the final occurrence of $A$ in $\alpha$. Form the string $\beta$ by deleting this final occurrence of $A$ and all the characters that follow it (the characters following it must all be $B$). 
It is clear that $\beta$ contains exactly $m$ A's and at most $n$ B's. Further, the strings $\alpha$ and $\beta$ are in one-to-one correspondence with each other. (We already described how to obtain $\beta$ from $\alpha$. To write down $\alpha$ given $\beta$, concatenate $\beta$ with an $A$ followed by enough $B$'s to make the total number of $B$'s exactly equal to $n$.) 
Thus the number of permutations with $m$ A's and at most $n$ B's is equal to the number of permutations with exactly $m+1$ A's and exactly $n$ B's, which is clearly $\binom{m+n+1}{m+1}$. 
