# $\sum_{k=1}^{z-1}\binom{k-1}{n-2}=\binom{z-1}{n-1}$

I'm trying to find out why this is true: $$\sum_{k=1}^{z-1}\binom{k-1}{n-2}=\binom{z-1}{n-1}$$

In the solutions of my teacher, he doesn't explain anything more then this, so I guess it is something trivial, but I don't see why.

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From Pascal theorem $$\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$$ follow that $$\binom{m-1}{k-1}=\binom{m}{k}-\binom{m-1}{k}$$ using last equation we can write $$\sum_{k=1}^{z-1}\binom{k-1}{n-2}=\sum_{k=1}^{z-1}\Bigg(\binom{k}{n-1}-\binom{k-1}{n-1}\Bigg)=$$ $$=\binom{1}{n-1}-\binom{0}{n-1}+\binom{2}{n-1}-\binom{1}{n-1}+...+\binom{z-1}{n-1}-\binom{z-2}{n-1}=\binom{z-1}{n-1}$$ NOTE This relation is called Hockey stick theorem
Given $z-1$ different objects,
${z-1} \choose {n-1}$ is the number of ways to select a subset of $n-1$ of them.
Suppose the highest-numbered object chosen is number $k$. After selecting this item, you still have $n-2$ objects to choose, and they can be any of the first $k-1$ objects, so there are ${k-1} \choose {n-2}$ ways to do this.