I need to prove the following: ${n\choose m-1}+{n\choose m}={n+1\choose m}$, $1\leq m\leq n$.
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 not really used to calculations with factorials and can't make much sense from it…
Best Answer
This is the most simplest answer,
$$\begin{align*}\begin{split} {n\choose m-1}+{n\choose m} &= \frac{m}{m}\cdot\frac{n!}{(m-1)!(n-m+1)!}+\frac{(n+1-m)}{(n+1-m)}\cdot\frac{n!}{m!(n-m)!}\\ &=\frac{mn!}{(m)!(n-m+1)!}+\frac{(n+1-m)n!}{m!(n+1-m)!} \\ &=\frac{mn!+(n+1)n!-mn!}{(m)!(n-m+1)!}\\ &=\frac{(n+1)n!}{(m)!(n-m+1)!} \\ &=\frac{(n+1)!}{(m)!(n-m+1)!} ={n+1\choose m}\end{split}\end{align*}$$