TITLE: binomial coefficient equal to sum
QUESTION [6 upvotes]: I have observed the following relation
$$\binom{n}{k}=\sum\limits_{i=0}^{k} \binom{n-i-1}{k-i}, \quad \frac{n}2\leq k \leq n-1$$
Numerical example with $n=5$ and $k=3$
$$\binom{\color{red}5}{\color{blueviolet}3}= \binom{\color{Magenta}4}{\color{blueviolet}3}+\binom{\color{blueviolet}3}{\color{Orange}2}+\binom{\color{Orange}2}{\color{YellowGreen}1}+\binom{\color{YellowGreen}1}{\color{Brown}0}$$
$$10\  \ \  = \  \ \ 4 \ \  \ +\ \  \ 3 \ \  \ +  \  \ \  2  \  \  \ \ + \ \ 1$$
Is this relation true? If yes, is there a name for it?
I´ve used identities like $\binom{n+1}{k+1} = \binom nk + \binom n{k+1}$ to show the relation above, but I failed. Does anyone have an idea how to manage it?
Thanks for taking time to read the question.

REPLY [6 votes]: This can also be shown with negative binomial coefficients and Vandermonde's Identity:
$$
\begin{align}
\sum_{i=0}^k\binom{n-i-1}{k-i}
&=\sum_i\binom{n-i-1}{k-i}\binom{i}{i}\tag1\\
&=\sum_i\binom{k-n}{k-i}(-1)^{k-i}\binom{-1}{i}(-1)^i\tag2\\
&=(-1)^k\binom{k-n-1}{k}\tag3\\[3pt]
&=\binom{n}{k}\tag4
\end{align}
$$
Explanation:
$(1)$: remove the limits from the summation
$(2)$: negative binomial coefficients
$(3)$: Vandermonde Identity
$(4)$: negative binomial coefficients