Prove with a combinatorial proof (story): $\displaystyle\sum_{k=0}^n k \binom n k ^2=n\binom{2n-1}{n-1}$
My attempt:
Let's make it easier to work with (it's very easy to show that identity): $\displaystyle\sum_{k=0}^n \frac kn \binom n k ^2=\binom{2n-1}{n-1}\Rightarrow \sum_{k=0}^n \binom {n-1}{k-1} \binom n k =\binom{2n-1}{n-1}$
RHS: we want to form a group of size $n-1$ from $2n-1$ people.
LHS: The group has $n$ men and $n-1$ women. We'll choose the men and women such that there will always be $1$ men more than women:
No men - no way to form such group.
$1$ person - $n=\binom {n-1} 0 \binom {n} 1$.
$3$ people - $\binom{n-1} 1\binom n 2$
...
$n$ people - $\binom{n-1} {n-1}\binom n n$
And if we sum up all the different cases we get: $\displaystyle\sum_{k=0}^n \binom {n-1}{k-1} \binom n k$
I'm not convinced with my story for the LHS, picking the groups such that there will always be $1$ men more seems too arbitrary. Also I'm pretty sure the first and last cases are wrong. Any tips on how fix it?
Note: no integrals nor use of other identities without proving them nor generating functions.
Left-hand side: Pick $k$ women to include, and $k$ men to exclude, to form a team of $n$ people. Choose one of the $k$ women to lead.
Right-hand side: Pick one of the $n$ women to lead. Pick $n-1$ of the other $2n-1$ people to form the team.