Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then $\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$.

Question

I'm reading Ghorpade's A Course in Calculus and Analysis.

Given $a\in \mathbb{R}$ and $k\in \mathbb{Z}$, the binomial coefficient associated with $a$ and $k$ is defined by:

$$\binom ak = \begin{cases} \frac{a(a-1)\dots(a-k+1)}{k!} & \text{if }\ k \geq 0 \\ 0 & \text{if }\ k < 0 \end{cases}.$$

Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then:

$$\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$$

I guess I've seen the second version of the binomial theorem, but I don't know how to prove the second through the first. I really have no clue here.

Answer 1

$$\begin{eqnarray*} {\binom ak}&=&{\frac{a!}{k!(a-k)!}} \\ {}&&{}\\ {\binom {a}{a-k}}&=&{\frac{a!}{(a-k)!(a-[a-k])!}} \\ {}&&{}\\ {}&=&{\frac{a!}{(a-k)!(a-a+k)!}} \\ {}&&{}\\ {}&=&{\frac{a!}{(a-k)!k!}} \\ {}&&{}\\ {}&=&{\frac{a!}{(a-k)!k!}} {}&&{}\\ {}&&{}\\ {\frac{a!}{k!(a-k)!}}&=&{\frac{a!}{(a-k)!k!}}\implies \binom ak=\binom {a}{a-k} \end{eqnarray*}$$

Answer 2

This is kind of strange since I thought you normally started with the second and then went to the first. Anyway, just multiply top and bottom of first definition by $(a-k)!$.