I would like help with combinatorial proof , not algebraic proof . Thank you for your time
$\sum_{i=0}^{n}(-1)^i\binom{n}{i}\binom{n-i+k-1}{k}=\binom{k-1}{k-n}$
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Sign up to join this communityI would like help with combinatorial proof , not algebraic proof . Thank you for your time
$\sum_{i=0}^{n}(-1)^i\binom{n}{i}\binom{n-i+k-1}{k}=\binom{k-1}{k-n}$
Denote $[n]=\{1,\ldots,n\}$.
We choose an $i$-subset $A\subset [n]$, then a $k$-multiset $B\subset [n]\setminus A$, and take $(-1)^i$ for each such a choice. Note that if $B$ is fixed, the sum of $(-1)^{|A|}$ over subsets $A\subset [n]\setminus B$ equals to 0 unless $B=[n]$, when the sum equals 1. Therefore we should count $k$-multisets $B$ which cover $[n]$. These are in bijection with $(k-n)$-multisets of $[n]$, there exist exactly ${k-n+(n-1)\choose k-n}={k-1\choose k-n}$ such multisets.