Take $\binom{n}{r}$. It denotes how in how many different ways you can choose $r$ elements from a set of $k$ elements. For case $\binom{4}{3}$ which evaluates to $\frac{4!}{3!(4-3)!}=4$, it perfectly makes sense.
However, consider $\binom{-4}{3}$. Evaluating it by factoring out $(n-r)!$ from the numerator and the denominator, we get $\frac{(-4)_{3}}{3!}$, where $(-4)_{3}$ is a falling factorial. Hence, $\binom{-4}{3}=-20$.
Now, how do we explain this result? Following the combinatoric reasoning, how can there be a negative number of ways to choose $r$ elements from a set containing a negative number of elements? And if counting with negative ways and negative elements is possible, why can't we check how many negative ways there are of choosing $r$ elements, for example $\binom{4}{-3}$?