TITLE: Help finding a combinatorial proof of $k {n \choose k } = n {n - 1 \choose k -1}$
QUESTION [11 upvotes]: Help finding a combinatorial proof of $k {n \choose k } = n {n - 1 \choose k -1}$
I have expanded it this far:
$$\frac{k \cdot n!}{k!(n-k)!} = \frac{n \cdot (n-1)!}{(k-1)!(n-k)!} $$
but then I am stuck from where to go from there.

REPLY [3 votes]: Left Side =
 $k\binom{n}{k} = k\frac{n!}{k!(n-k)!}$  
Because $\frac{k}{k!} = \frac{1}{(k-1)!}$
$k\binom{n}{k}  = \frac{n!}{(k-1)!(n-k)!}$
Right side = 
 $n\binom{n-1}{k-1} = \frac{n(n-1)!}{(k-1)!(n-1-k+1)!} = \frac{n(n-1)!}{(k-1)!(n-k)!}$  
Because $n(n-1)! = n!$
$n\binom{n-1}{k-1} = \frac{n!}{(k-1)!(n-k)!}$
Left Side = Right side, therefore proved !!