I need to prove the following: ${n\choose m}={n\choose n-m}$
With the definition: ${n\choose m}= \left\{ \begin{array}{ll} \frac{n!}{m!(n-m)!} & \textrm{für \(m\leq n\)} \\ 0 & \textrm{für \(m>n\)} \end{array} \right.$
and $n,m\in\mathbb{N}$.
I'm stuck at how to even start this. Using induction? Any help would be appreciated, I tried to search SA, but couldn't find an answer.
An other way:
to choose $m$ balls in a box that contain $n$ balls is equivalent that to not choose the $n-m$ other balls.
Hint: Just use factorial formula.
$${n\choose m}= \frac{n!}{m! \cdot (n-m)!}$$
Solution:
$${n\choose n-m} = \frac{n!}{(n-m)! \cdot (n -(n-m))!} = \frac{n!}{(n-m)! \cdot m!} = \frac{n!}{m! \cdot (n-m)!} = {n\choose m}$$
I believe, rest is obvious.
$$ {n\choose n-m} = \frac{n!}{(n-m)!(n-(n-m))!} = \frac{n!}{(n-m)!m!} = {n \choose m} $$