Binomial Coefficient with Zero

Theorem

 * $\displaystyle \forall r \in \R: \binom r 0 = 1$

The usual presentation of this result is:
 * $\displaystyle \forall n \in \N: \binom n 0 = 1$

Proof
From the definition of binomial coefficients:


 * $\displaystyle \binom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$

where $r^{\underline k}$ is the falling factorial.

In turn:


 * $\displaystyle x^{\underline k} := \prod_{j \mathop = 0}^{k-1} \left({x - j}\right)$

But when $k = 0$, we have:
 * $\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right) = 1$

as $\displaystyle \prod_{j \mathop = 0}^{-1} \left({x - j}\right)$ is a vacuous product.

From the definition of the factorial we have that $0! = 1$.

Thus:


 * $\displaystyle \forall r \in \R: \binom r 0 = 1$

This is completely compatible with the result for natural numbers:
 * $\displaystyle \forall n \in \N: \binom n 0 = 1$

as from the definition:
 * $\displaystyle \binom n 0 = \dfrac {n!} {0! \ n!}$

the result following directly, again from the definition of the factorial where $0! = 1$.

Also see

 * Particular Values of Binomial Coefficients for other similar results.