TITLE: Is $n \choose k$ defined when $k < 0$? What about $n < k$?
QUESTION [12 upvotes]: 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 $k < 0$? What about $n < k$? And in either case, what's the reasoning?

REPLY [2 votes]: 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 $x=0,1,\dots,k-1$.