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I want to prove the following identity combinatorial: \begin{align} {n + k \choose k} = \sum \limits_{i=0}^n {n -i +k -1 \choose k -1}. \end{align} We want to choose k out of $n+k$.
I want to use the following identity: \begin{align} {n \choose k} = {n \choose n-k}. \end{align} So we have to prove that: \begin{align} {n + k \choose n} = \sum \limits_{i=0}^n {n -i +k -1 \choose n -i}. \end{align} In words, choose n out of n+k. I still can not figure out why this is true. Can some body help me out?

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We have $n+k$ different doughnuts, lined up in a row. They are labelled $D_0$ to $D_{n+k-1}$. The last $k$ are chocolate covered. There are $\binom{n+k}{k}$ ways to choose $k$ doughnuts for breakfast. Let us count the number of ways to do the choosing in another way.

Let $i$ be the label of the leftmost doughnut chosen. Then $i$ goes from $0$ to $n$. For any given $i$, that leaves $n-i+k-1$ doughnuts, from which we must choose $k-1$. So for any $i$, there are $\binom{n-i+k-1}{k-1}$ ways to do the choosing. For all the ways to choose breakfast, add up from $i=0$ to $i=n$.

Remark: We could reword the example to use the identity $\binom{n+k}{k}=\binom{n+k}{n}$ mentioned in the question, but that is unnecessary.

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