How to prove $\binom{-r}{k} = (-1)^k \binom{r + k - 1}{k}$?

Question

I encountered the following problem in Brualdi (Section 5.6, Problem 21):

Prove that, for all real numbers $r$ and all integers $k$,

$$\binom{-r}{k} = (-1)^k \binom{r + k - 1}{k}$$

It looks like the right hand side has the formula for choosing $k$ elements from a multiset.

$$\binom{r + k - 1}{k} = \binom{r + k -1}{k - 1}$$

I know that $\binom{n}{k} = 0$ if $n < k$, but what if $k$ is negative and $n$ is a real number. I'm not sure where to go from here.

Answer 1

$$\binom{-r}{k} =$$

$$ \frac {(-r)(-r-1)(-r-2)...(-r-k+1)}{k!}=$$

$$(-1)^k \frac {(r+k-1)(r+k-2)...(r+1)(r)}{k!}=$$

$$(-1)^k\binom{r+k-1}{k}$$

Answer 2

\begin{eqnarray*} \binom{n}{k}=\frac{\overbrace{n(n-1)\cdots(n-k+1)}^{k \text{terms}} }{k!} \end{eqnarray*} So \begin{eqnarray*} \binom{-r}{k}=\frac{(-r)(-r-1)\cdots(-r-k+1)}{k!}=(-1)^{k} \frac{(r+k-1)!}{k!(r-1)!} =(-1)^{k}\binom{r+k-1}{k}. \end{eqnarray*}

Answer 3

The starting definition of the binomial coefficients, for general complex argument values, is:

$${n \choose k} \equiv \frac{\Gamma (n+1)}{\Gamma (k+1) \Gamma (n-k+1)}.$$

(Note that this ratio is undefined for non-positive integers $n$ or $k$.) Using this formula we have:

$$\begin{equation} \begin{aligned} {-r \choose k} =\frac{\Gamma (1-r)}{\Gamma (k+1) \Gamma (1-r-k)} &= \frac{\Gamma (1-r)}{\Gamma (k+1) \Gamma (1-r-k)} \cdot \frac{\Gamma (r)}{\Gamma (r)} \cdot \frac{\Gamma (r+k)}{\Gamma (r+k)} \\[8pt] &= \frac{\Gamma (r+k)}{\Gamma (k+1) \Gamma (r)} \cdot \frac{\Gamma (1-r) \Gamma (r)}{\Gamma (1-r-k) \Gamma (r+k)} \\[8pt] &= \frac{\Gamma (r+k)}{\Gamma (k+1) \Gamma (r)} \cdot \frac{\pi}{\sin (\pi r)} \Bigg/ \frac{\pi}{\sin (\pi (r+k))} \\[8pt] &= \frac{\Gamma (r+k)}{\Gamma (k+1) \Gamma (r)} \cdot \frac{\sin (\pi (r+k))}{\sin (\pi r)} \\[8pt] &= \frac{\Gamma (r+k)}{\Gamma (k+1) \Gamma (r)} \cdot (-1)^k \\[8pt] &= (-1)^k {r+k-1 \choose k}. \end{aligned} \end{equation}$$

(The third line of this working is using Euler's reflection formula for the gamma function.) This completes the proof for all cases where the gamma function is well-defined. The remaining cases involving $r = 0, 1, 2, ...$ and $k = 0, -1, -2, ...$ are generally resolved by extending the definition of the binomial coefficients using this very formula (in which case the formula holds by definition).

This paper on the binomial coefficients with negative arguments might also assist you.