Prove using combinatorics $\sum\limits_{i=k}^{n-r+k}\binom{i}{k}\binom{n-i}{r-k}=\binom{n+1}{r+1}$

Question

Prove using combinatorics $\sum\limits_{i=k}^{n-r+k}\binom{i}{k}\binom{n-i}{r-k}=\binom{n+1}{r+1}$.

The right side is sth clear choosing $r+1$ balls from $n+1$ different balls.The left side want to choose $r$ balls from $n$ balls but we should consider $i$ black balls and $n-i$ white balls then choose $k$ black and $r-k$ white balls.Because the main work in left side is choosing $r$ balls then I used to do the proof like the proof of $\sum\limits_{i=r}^n\binom{i}{r}=\binom{n+1}{r+1}$but it didn't help me.Any hints?

"Different balls" makes it harder than "balls labeled $1,2,...,n+1$." Given $r+1$ balls taken from this list, sort them, and let $i+1$ be the $k+1$st ball, in order. — Thomas Andrews, Apr 03 '17 at 16:36

score 1 · Accepted Answer · answered Apr 03 '17 at 16:38

Note: For the equation to make sense, we need $0\le k\le r$.

We can pick $r+1$ numbers $a_0<a_1<\ldots <a_k<\ldots <a_r$ from the $n+1$ numbers $0,1,2,\ldots, n$ in $n+1\choose r+1$ ways.

In any such choice, the $(k+1)$st smallest number $a_k$ will fulfill the inequality $k\le a_k\le n-r+k$. Then for $a_0,\ldots,a_{k-1}$ there are $a_k\choose k$ choices and for $a_{k+1},\ldots, a_r$, there are $n-a_k\choose n-k$ choices. Thus in total there are $$ \sum_{a_k=k}^{n-r+k}{a_k\choose k}{n-a_k\choose n-k}$$ possibilities. Writing $i$ instead of $a_k$ yields the desired result.

Prove using combinatorics $\sum\limits_{i=k}^{n-r+k}\binom{i}{k}\binom{n-i}{r-k}=\binom{n+1}{r+1}$

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