Finding an algebraic proof for $r{n \choose r} = n{n-1 \choose r-1}$ I can't seem figure this proof out.
How are both sides equal.
$$r{n \choose r} = n{n-1 \choose r-1}$$

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