It's easy to see that $ \sum_i {A\choose i} {B\choose n-i} = {A+B\choose n} $ since when we choose $n$ things out of $A+B$, some ($i$ of them) are in the $A$ and the rest are in the $B$.
Is there any reasonable formula for $ \sum_{i< I} {A\choose i} {B\choose n-i} = {A+B\choose n}, $ i.e. we have a bound on how many of them are from the $A$ side?
Is the $B=1$ case any easier? (That being my real question.)
EDIT: I totally misasked this question, and have hopefully fixed it here: Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$?