Partial sum of ${A \choose i} {B\choose ni}$, when $B=1$?
It is very easy to see that $$ \sum_i {A\choose i} {B\choose ni} = {A+B\choose n} $$ given that when we pick $n$ points out of $A+B$, some ($i$ of them) remain in the $A$ et cetera remain in the $B$.

Exists any kind of practical formula for $$ \sum_{i< I} {A\choose i} {B\choose ni} = {A+B\choose n}, $$ i.e. we have a bound on the amount of of them are from the $A$ side?

Is the $B=1$ instance any kind of less complicated? (That being my actual inquiry.)
EDIT: I entirely misasked this inquiry, and also have actually with any luck repaired it below: Partial sum over $M$, of ${m+j \choose M} {1M \choose m+iM}$?
For B = 1 this is instead very easy:
${1 \choose j}$ is $1$ if $j$ is $0$ or $1$, and also is $0$ for any kind of various other value of $j$. So your first expression comes to be $ {A\choose n1} {1\choose 1} + {A\choose n} {1\choose 0} = {A+1 \choose n} $ or as Fabian claims, Pascal is regulation
$$ {A\choose n1} + {A\choose n} = {A+1 \choose n}. $$
Your 2nd expression relies on $I$. If $I \le n1$ after that it offers an amount of 0 ; if $I = n$ and also $i \lt I$ it offer an amount of ${A\choose n1}$, and also if $I \gt n$ it offers an amount of ${A+1 \choose n}$.
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