Help finding a combinatorial proof of $k {n \choose k } = n {n - 1 \choose k -1}$ 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.
 A: Here is a more comprehensive combinatorial proof.
Given $n$ people, we can form a committee of size $k$ in ${n\choose k}$ ways. Once the committee is formed we can pick a committee leader in $k$ ways. Thus we can form a committee of size $k$ with a committee leader in $k{n\choose k}$ ways. We can count the same thing by first picking a committee leader in $n$ ways. Once this has been done we can form the committee with the remaining $n-1$ people in ${n-1\choose k-1}$ ways. We choose $k-1$ people from $n-1$ people because a leader has already been specified. Thus we can pick a committee leader first and then form the committee with the remaining people in $n{n-1\choose k-1}$ ways. Hence $k{n\choose k}=n{n-1\choose k-1}$.
A: 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 !! 
A: To select a committee of $k$ people with a president from a group of $n$ people, you can


*

*select the $k$ committee members, and then select the president from the members, or

*select the president from the $n$ people, and the $k-1$ remaining committee members from the remaining $n-1$ people.

