How to prove the equation $\sum^{N-r}_{x=0}\binom{N-x}r =\binom{N+1}{r+1}$

Question

I was trying to solve a probabilistic problem and I found I had to use the equality $\displaystyle\sum^{N-r}_{x=0}\begin{pmatrix} N-x \\ r \end{pmatrix}=\begin{pmatrix} N+1 \\ r+1 \end{pmatrix}$ in an intermediate step. I verified it with Wolfram Alpha, but I really can't come up with a proof (I tried expanding everything out but find no clue in the end). I hope I can find a proof here and it is even more delighting if someone is willing to teach me some typical ways of proving equations involving combinations and permutations, which seem to be unmentioned in the textbook I am currently using.

try with "common properties of Pascal's Triangle" – Exodd Mar 16 '18 at 17:28 — Exodd, Mar 16 '18 at 17:28

score 1 · Accepted Answer · answered Mar 16 '18 at 17:32

1

That's known as the Hockey-stick identity (from Pascal's Triangle trick)

$$\sum^n_{i=r}{i\choose r}={n+1\choose r+1} \qquad \text{ for } n,r\in\mathbb{N}, \quad n>r$$

answered Mar 16 '18 at 17:32

user

154,566

Oh, thank you very much. I never heard about that identity before, so I couldn't know what to search on the Internet. – ZhangTianrong Mar 16 '18 at 17:47
@ZhangTianrong You are welcome! Bye – user Mar 16 '18 at 17:49

How to prove the equation $\sum^{N-r}_{x=0}\binom{N-x}r =\binom{N+1}{r+1}$

1 Answers1