what happens when in a combination, $n \choose k$, $n < k$? Is this 0? What happens when in a combination, $n \choose k$, $n &lt; k$? Does this result in a zero, or is this undefined?
Not necessarily $0$. We have definition
$$\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$$
So for example if we take $n = -3$ and $k = 2$, we have
$$\binom{-3}{2} = \frac{(-3)\cdot(-4)}{2!} = 6$$
If you don't believe, see https://www.wolframalpha.com/input/?i=C(-3,2). Selection of objects is just an analogy for natural number values of $n$ and $k$, I suggest you to use the definition instead, for a more general result.
We have $n \choose r$. (The common notation is $n$ and $r$)

\begin{align}
n \choose r &amp; = \frac{n!}{(n-r)!r!}\\
\end{align}

$n \lt r \implies n-r &lt; 0$. The factorial is not defined for negative numbers, so $n \choose r$ does not have an answer if $n &lt; r$.