Clarification of a proof of an identity for binomial coefficients, $\binom{n}{k}= \binom{n-1}{k-1}+\binom{n-1}{k}$

Question

I am studying analysis 1 (first term). So there is this definition in our textbook:

for every $n \in \mathbb{N} \ge 1$ and every k $\in \mathbb{Z}$ there is

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

This definition is then proved for $0<k<n$ : (This proof is 100% correct, i guess.) It starts with:

$\binom{n-1}{k-1}+\binom{n-1}{k}=\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}$

I understand the rest of the proof, its just this transformation that is driving me nuts. The right side makes sense for me. But for the left side, doesn't it have to be

$\binom{n-1}{k-1}+\binom{n-1}{k}=\frac{(n-1)!}{(k-1)!(n-1-k-1)!}+\frac{(n-1)!}{k!(n-k-1)!}$

When we use the formula

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

? Why can he just leave out the -2? Or am i just completely blind.

Answer 1

You forgot the parenthesis in the denominator. It should be $(n-1)-(k-1)$ which gives $n-k$.

You wrote $n-1-k-1$ which is how you got the incorrect $n-k-2$.

Answer 2

Actually it's not a definition. Anyway, an other way to prove it:

$\binom{n}{k}$ is the number of subset of $k$ element of a set with $n$ elements. Let $a\in E$ where $E$ is a set with $n$ elements. To count the subset of $E$ with $k$ element you can count the subset that contain $a$ (and they are $\binom{n-1}{k-1}$) and the subset that doesn't contain $a$ (an they are $\binom{n-1}{k}$), you finally get $$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}.$$