Is $n \choose k$ defined when $k < 0$? What about $n < k$?

It is customary to define $\binom{n}k=0$ for $0\le n<k$. 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 $n<k$. 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 $k<0$ as well, of course); this is convenient in some induction proofs, for instance.

You can use the gamma function to extrapolate the factorial function to all real (or complex) numbers except negative integers, but this involves analysis and is kind of arbitrary. The binomial coefficient $\binom x k$ is naturally defined for all integers $k\ge0$ and all real or complex numbers $x$, as it is simply a polynomial function of $x$, namely $x(x-1)(x-2)\cdots(x-k+1)/k!$. Of course the value is $0$ for $x=0,1,\dots,k-1$.