How can I prove that $\left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| = \binom{n}{r}$, only for $a=0$ and $a=n$?

Question

In a previous question I asked about the maximum module reached by the quantity $f_{n,r} (a) = \sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}$. Now I ask when this maximum value can be reached.

This is a conjecture:

How can I prove that the equality

\begin{equation} \left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| = \binom{n}{r} \end{equation}

where $0\leq a \leq n$, $0\leq r \leq n$ and $n,r,a \in \mathbb{N}$, $\mathbf{r \neq n, r\neq0}$, holds only for $a=0$ and for $a=n$?

What I'm trying to prove is that for $r$ even I cannot have, for any $0\leq a \leq n$ with $r \neq n, r\neq0$:

\begin{equation} \sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i} = - \binom{n}{r} \end{equation}

Answer 1

This conjecture is false. Let $n = 2$, $a = 1$, and $r = 2$. The series becomes:

$$ {{1}\choose{0}} {{1}\choose{2}} + - {{1}\choose{1}} {{1}\choose{1}} + {{1}\choose{2}} {{1}\choose{0}} = 0 - 1 + 0 = -1 = - {{2}\choose{2}} = - {{n}\choose{r}} $$

Answer 2

Since

$$\sum_{i=0}^r\binom{a}i\binom{n-a}{r-i}=\binom{n}r\;,$$

the equality

$$\left|\,\sum_{i=0}^r(-1)^i\binom{a}i\binom{n-a}{r-i}\,\right|=\binom{n}r$$

holds if and only if the summation has exactly one non-zero term. This is the case precisely when there is a unique $i\in\{0,\ldots,r\}$ such that $i\le a$ and $r-i\le n-a$, i.e., such that

$$r+a-n\le i\le a\;.$$

This is the case if and only if $r+a-n=a$, or $n=r$.