Partial sum of ${A \choose i} {B\choose n-i}$, when $B=-1$?

It is very easy to see that $$ \sum_i {A\choose i} {B\choose n-i} = {A+B\choose n} $$ 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 $$ \sum_{i< I} {A\choose i} {B\choose n-i} = {A+B\choose n}, $$ 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 ${m+j \choose M} {1-M \choose m+i-M}$?

4
2022-06-07 14:33:36
Source Share
Answers: 1

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 n-1} {1\choose 1} + {A\choose n} {1\choose 0} = {A+1 \choose n} $ or as Fabian claims, Pascal is regulation

$$ {A\choose n-1} + {A\choose n} = {A+1 \choose n}. $$

Your 2nd expression relies on $I$. If $I \le n-1$ after that it offers an amount of 0 ; if $I = n$ and also $i \lt I$ it offer an amount of ${A\choose n-1}$, and also if $I \gt n$ it offers an amount of ${A+1 \choose n}$.

1
2022-06-07 15:05:33
Source