combinatoric proof $\sum_{k=1}^{n} k{n \choose k}^2=n{2n-1 \choose n-1}$

Question

$\sum_{k=1}^{n} k{n \choose k}^2=n{2n-1 \choose n-1}$

my attempt:

(in the first note that :$n{2n-1 \choose n-1}=n{2n-1 \choose n}$

$\sum_{k=1}^{n} k{n \choose k}^2=\sum_{k=1}^{n} k{n \choose k}{n \choose n-k}=0{n \choose 0}{n \choose n-0}+1{n \choose 1}{n \choose n-1}+2{n \choose 2}{n \choose n-2}+......+(n-1){n \choose n-1}{n \choose 1}+n{n \choose n}{n \choose 0}$

let's assume that we have two set A and B that have n element for each,such that $A \cap B =\emptyset$

so $|A|=n$ and $|B|=n$

$0{n \choose 0}{n \choose n-0} $ it is the number of ways to choose 0 element from A and n element from B 0 time .

$1{n \choose 1}{n \choose n-1} $ it is the number of ways to choose 1 element from A and n-1 element from B 1 time .

$2{n \choose 2}{n \choose n-2} $ it is the number of ways to choose 2 element from A and n-2 element from B 2 time .

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$n{n \choose n}{n \choose n-n} $ it is the number of ways to choose n element from A and 0 element from B n time .

so thier sum is the number of ways to choose n element from A and B n time .

and we know that $|A+B|=|A+|B|=2n$

. so it equal $n{2n \choose n} $

i know that's wrong but maybe my attempt can be devloped ...so now i search for my mistake

Answer 1

Suppose we have $n$ women and $n$ men (for a total of $2n$ people) and we wish to make a committee with a female president out of these people of size $n$.

We can do this in a few different ways:

• First, pick the female president in $n$ possible ways. Then, among the remaining $2n-1$ people, choose $n-1$ of them to serve as regular members on the committee.

$$n\binom{2n-1}{n-1}$$

• Break apart into cases based on how many women end up on the committee, letting the number of women in each case be called $k$. With $k$ women on the committee, choose which $k$ of the $n$ women they happen to be. Then, choose which of those $k$ women were designated as being president. Finally, of the $n$ men choose $k$ of the men to not be on the committee (leaving the remaining $n-k$ men to serve on the committee)

$$\sum\limits_{k=1}^nk\binom{n}{k}^2$$

As these expressions both are valid ways of counting the same scenario, they must be equal. QED