Is $n \choose k$ defined when $k < 0$? What about $n < k$? I know that ${n \choose 0} = 1$, and this makes sense to me based on my understanding of combinatorics.

But what about ${n \choose -1}$? My instinct is that this is undefined, since it is equivalent to $\frac{n!}{k!(n - k)!}$, which when $k = -1$ equals $\frac{n!}{(-1)!(n + 1)!}$, and factorials are only defined for non-negative integers.

Similarly, ${5 \choose 6} = \frac{5!}{6!(5 - 6)!} = \frac{5!}{6!(-1)!}$ should also be undefined. However, intuitively it seems to me that "5 choose 6" could also equal $0$, since there are no combinations of 6 objects to be made from a set of only 5 objects.

So, is $n \choose k$ defined when $k &lt; 0$? What about $n &lt; k$? And in either case, what's the reasoning?
It is customary to define $\binom{n}k=0$ for $0\le n&lt;k$. If you restrict $n$ and $k$ to the non-negative integers and think of $\binom{n}k$ as the number of $k$-element subsets of an $n$-element set, then $\binom{n}k$ should certainly be $0$ when $n&lt;k$. If instead you use the more general algebraic definition of the binomial coefficient,

$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}$$ 

for real $x$ and non-negative integer $k$, you automatically get $\binom{n}k=0$ when $n$ is a non-negative integer less than $k$. (Here $x^{\underline k}$ is a falling factorial.)

Not everyone defines $\binom{n}k$ for negative integers $k$, but when it is defined for them, it’s defined to be $0$. This ensures that the Pascal triangle identity,

$$\binom{n}k=\binom{n-1}{k-1}+\binom{n-1}k\;,$$ 

holds even for $k=0$ (and for integers $k&lt;0$ as well, of course); this is convenient in some induction proofs, for instance.
As one of the other users suggested, you can indeed extend $\binom{n}{k}$ to cases where $k&lt;0$ using the Gamma function. However, there are more ways to do it. For example, if we want to have $n$ be negative. Let $r=|n|$ and $f=r+k-1$, then

\begin{align}\binom{-r}{k} &amp;= \frac{-r\cdot-(r+1)\dots-(r+k-2)\cdot-(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&amp;=(-1)^k\;\frac{r\cdot(r+1)\cdot(r+2)\cdots(f-2)\cdot(f-1)\cdot f}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&amp;=(-1)^k\binom{f}{k}\\
&amp;=(-1)^k\left(\!\!\binom{f-k+1}{k}\!\!\right)\\
&amp;=(-1)^k\left(\!\!\binom{r}{k}\!\!\right)\;,\end{align}
where $\left(\!\!\binom{r}{k}\!\!\right)=\frac{r^{(k)}}{k!}=\frac{r\cdot(r+1)\cdot(r+2)\cdots(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}$. These types of things arise naturally in the study of multisets. In fact, one uses these often in Mathematical Biology while studying genetics. However, this is beyond "normal" probability/combinatorics. The reason it is often the case that $0\leq k$ and $0 \leq n$ is $k$ and $n$ often represent a number of objects, which must be nonnegative. The reason that $k\leq n$ is just as you suggest, one would obtain negative factorials (which require the Gamma function). Moreover, in the case where $2=n&lt;k=3$, how does one take 3 total objects from a possible 2? It can't be done nor does it make sense. So it is natural to only consider $k\leq n$.