What is zero choose one? Algebraically it comes out to be undefined- but if I have zero elements, and I'm asked to pull elements from it, this should just be zero, right?
 A: Combinatorically speaking, you are absolutely right. There is no way to choose $1$ element from an empty set, so $\binom01=0$. It's a good idea to keep in mind that we do not always have $\binom{n}{k}=\frac{n!}{k!(n-k)!}.$ That formula only holds for integers $n,k$ such that $0\le k\le n$. If $n,k$ do not meet all those conditions, then $\binom{n}{k}=0.$
A: One way to see this is with the binomial expansion,
$$
(1+x)^{\alpha}=\sum_{k=0}^{\infty}{{\alpha}\choose{k}}x^k.
$$
Since $(1+x)^{0}=1$, we must have ${{0}\choose{0}}=1$ and ${{0}\choose{k}}=0$ for $k>0$.
A: There is an interpretation of ${x \choose k}$ for positive integer $k$ but general $x$ as $$\frac{x \times (x-1) \times \cdots \times (x-k+1)}{k!}$$ which would give ${0 \choose 1}=0$.  
Indeed it gives ${0 \choose k}=0$ for positive integer $k$, and ${n \choose k}=0$ for positive integers $n$ and $k$ with $n \lt k$. This corresponds to the combinatorial interpretation.
In this interpretation ${x \choose 0}$ is typically the empty product, namely $1$. This includes ${0 \choose 0}$.
What is different is that this interpretation of ${x \choose k}$ gives something non-zero for fractional or negative integer $x$.
A: The graphic of the $\Gamma$ or factorial function can be found here : notice how it diverges towards $\pm\infty$ for each negative integer. Indeed, $$n!=\frac{(n+1)!}{n+1}\iff(-1)!=\frac{(-1+1)!}{-1+1}=\frac{0!}{0}=\frac10=\pm\infty$$ $$C_0^1=\frac{0!}{1!(0-1)!}=\frac1{(-1)!}=\frac1{\pm\infty}=0$$
A: It is zero.  You cannot draw a sample of size 1 out of a set of size 0.  
