binomial coefficient equal to sum

See my comment on your question.


The hockeystick identity tells us that: $$\sum_{i=r}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$

Applying this we find:

$$\sum_{i=0}^{k}\binom{n-i-1}{k-i}=\sum_{i=n-1-k}^{n-1}\binom{i}{n-1-k}=\binom{n}{n-k}=\binom{n}{k}$$


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