Having the following definition:
$$\binom \alpha k = \frac{\alpha^{\underline k}}{k!} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-k+1)}{k(k-1)(k-2)\cdots 1}\tag{1}$$
Why $\bbox[1px,border:1px solid black]{{-1\choose 3}=-1}$? I know that $\bbox[1px,border:1px solid black]{{-1\choose 3}=-1}$ because of the following identity:
$${-n \choose r}=(-1)^r{n+r-1 \choose r}$$
Then I'd just need to compute:
$${-1 \choose 3}=(-1)^3{1+3-1 \choose 3}=-1\times1=-1$$
But upon using $(1)$, I'd have:
$${-1 \choose 3} = \frac{-1(-1-1)(-1-2)(-1-3+1)}{3(2)(1)}=\frac{-1(-2)(-3)(-3)}{6}=\frac{18}{6}=3$$
I don't know what I'm doing wrong, I suppose I should do $-1(-1-1)(-1-2)(-1-3+1)$ until I reach $k$ which is $3$.
You have one factor too many in the numerator: it should be
$$\binom{-1}3=\frac{(-1)^{\underline 3}}{3!}=\frac{(-1)(-1-1)(-1-2)}{3!}=\frac{-6}6=-1\;.$$