# combinatoric proof $\sum_{i=0}^{n}(-1)^i\binom{n}{i}\binom{n-i+k-1}{k}=\binom{k-1}{k-n}$ [closed]

I would like help with combinatorial proof , not algebraic proof . Thank you for your time

$$\sum_{i=0}^{n}(-1)^i\binom{n}{i}\binom{n-i+k-1}{k}=\binom{k-1}{k-n}$$

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.