Combinatorics identity question

I encountered this particular combinatorial identity :$$\begin{align}&1\dbinom{n-1}{r-1}+2\dbinom{n-2}{r-1}+\cdots+(n+1-r)\dbinom{r-1}{r-1}\\=&\dbinom{n}{r}+\dbinom{n-1}{r}+\cdots+\dbinom{r}{r}\\=& \dbinom{n+1}{r+1}\end{align}$$How does one figure it out?

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$\binom{n+1}{r+1}$ is of course the number of ways to choose $r+1$ numbers from the set $\{0,1,\ldots,n\}$. Now break those sets into categories according to the smallest number chosen. If $k$ is the smallest number chosen, the remaining $r$ numbers must come from the $(n-k)$-element set $\{k+1,\ldots,n\}$, so there are $\binom{n-k}r$ ways to choose them. The smallest number chosen can be anywhere from $0$ through $n-r$, so


Finally, categorize the $(r+1)$-element subsets according to their second-smallest elements. Suppose that the second-smallest element of a set is $k$. Then $r-1$ elements must be chosen from the $n-k$ elements in $\{k+1,\ldots,n\}$, and the smallest element must be chosen from the $k$ elements in $\{0,1,\ldots,k-1\}$; these choices can be made in a total of $k\binom{n-k}{r-1}$ ways. The second-smallest element can be anything from $1$ through $n-r+1$, so


This is what they call the Hockey-Stick Identity or the Chu-Shih-Chieh’s Identity as I have encountered it in the book Principle and Techniques in Combinatorics by Chen and Koh. You can read about it from here. 🙂