Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.