Struggle with binomial coefficient notations I am trying to get used to the binomial notation. The general forumla for it is:
$$\binom{k}{n}=\dfrac{k(k-1)(k-2)\cdots(k-n+1)}{k!}$$
So let take $\binom{k-4}{2}$, that is going to equal to \begin{align} & \frac{k(k-1)(k-2)\cdots(k-4-2+1)}{2!} \\[8pt]
= {} &\frac{k(k-1)(k-2)\cdots(k-5)}{2} \end{align}
Is this the correct result?
 A: We have that
$$\binom{k}{n}=\dfrac{k!}{n!(k-n)!}=\dfrac{k(k-1)(k-2)...(k-n+1)}{n!}$$
and then
$$\binom{k-4}{2}=\dfrac{(k-4)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)(k-6)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)}{2}$$
A: Compare
$$\binom{j}{2} = \frac{j(j-1)}{2}$$
and substitute $j=k-4$.
A: 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 $$\binom{a} {n} =\frac{a!} {n! (a-n)!} $$ 
A: Remember also that in the expansion of the binomial, there are as many factors upstairs as downstairs, as long as you include the factors equal to $1$.
