TITLE: $\sum_{k=1}^{z-1}\binom{k-1}{n-2}=\binom{z-1}{n-1}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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