Struggle with binomial coefficient notations

We have that

$$\binom{k}{n}=\dfrac{k!}{n!(k-n)!}=\dfrac{k(k-1)(k-2)...(k-n+1)}{n!}$$

and then

$$\binom{k-4}{2}=\dfrac{(k-4)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)(k-6)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)}{2}$$

Compare

$$\binom{j}{2} = \frac{j(j-1)}{2}$$

and substitute $j=k-4$.

The binomial coefficient $\binom{a} {n} $ is defined for all $a\in\mathbb {C} $ and all $n\in \mathbb {N} $ via $$\binom {a} {n} =\frac{a(a-1)(a-2)(a-3)\dots (a-(n-1))}{n!}$$ The numerator contains $n$ factors starting with $a$ and decreasing by $1$ as one moves from one factor to the next. The last factor thus is $a-(n-1)=a-n+1$. The denominator also follows same pattern with the first factor being $n$.

Also by convention we define $\binom{a} {0}=1$. So for $\binom{k-4}{2}$ your numerator should start start with $k-4$ and the next factor is $k-5$ and your stop because only two factors are needed. The denominator is $2\cdot 1$ and thus we have $$\binom{k-4}{2}=\frac{(k-4)(k-5)}{2}$$

When $a$ is also a positive integer then you can prove that $$\binom{a} {n} =\frac{a!} {n! (a-n)!} $$