Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Use counting arguments to prove these identities:
$$1. \quad\begin{pmatrix}
n \\
0\\
\end{pmatrix}^2 + \begin{pmatrix} n \\ 1\\ \end{pmatrix}^2 +\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 + \ldots+\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
$$2. \quad\ \begin{pmatrix}
n \\
1\\
\end{pmatrix}^2 + 2\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 +3\begin{pmatrix} n \\ 3\\ \end{pmatrix}^2 + \ldots+n\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \frac n2\begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
Use $(1+x)^{2n}$ to prove Identity $1$.
Use Identity $1$ to prove Identity $2$.

 A: Consider a group of $n$ males and $n$ females from which a committee of $n$ need to be chosen.
There are $n + 1$ possible cases:
Case $1$: $0$ males and $n$ females are chosen- $\binom{n}{0} \cdot \binom{n}{n} = \binom{n}{0}^2 ways$
Case $2$: $1$ male and $n -1$ females are chosen- $\binom{n}{1} \cdot \binom{n}{n - 1} = \binom{n}{1}^2 ways$
..
Case $n + 1$: $n$ males and $0$ females are chosen- $\binom{n}{n} \cdot \binom{n}{0} = \binom{n}{n}^2 ways$.
But using a more direct approach, the selection can be made using $\binom{2n}{n}$ ways. 
Thus, $\binom{2n}{n} = \sum_{k = 0}^{n} \binom{n}{k}^2$.

For verifying the result using the binomial theorem,


*

*What is the coefficient of $x^n$ in $(1 + x)^{2n}$?

*What is the coefficient of $x^n$ in $(1 + x)^n \cdot (1 + x)^n$?

*For the second question, you should be able to find the derivative for the identity involving the binomial expansion of $(1 + x)^{2n}$ and proceed in a similar manner.

A: Part 1:
$$\begin{align}
(1+x)^{2n}&=\sum_{m=0}^{2n}\binom {2n}mx^m\tag{1}\\
(1+x)^{2n}&=(1+x)^n(1+x)^n\\
&=\sum_{r=0}^n\binom nr x^r\sum_{s=0}^n\binom ns x^s\\
&=\sum_{r=0}^n\sum_{s=0}^n\binom nr\binom nsx^{r+s}\\
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\tag{2}\\
[x^m](2)=[x^m](1):\qquad 
\sum_{r=0}^m\binom nr\binom n{m-r}&=\binom {2n}m\\
\text{Put }m=n:\qquad
\sum_{r=0}^n\binom nr\binom n{n-r}&=\binom {2n}n\\
\sum_{r=0}^n\binom nr^2&=\binom {2n}n\color{red}\blacksquare
\end{align}$$
Part 2:
$$\begin{align}
(1)=(2):\qquad 
\sum_{m=0}^{2n}\binom {2n}mx^m&=\sum_{m=0|}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\\
\frac d{dx}:\qquad 
\sum_{m=0}^{2n}\binom {2n}m mx^{m-1}
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}mx^{m-1}\\
[x^{m-1}]:\qquad \qquad
m\binom {2n}m&=\sum_{r=0}^m m\binom nr\binom n{m-r}\\
\text{Put }m=n:\qquad 
n\binom {2n}n
&=\sum_{r=0}^n n\binom nr\binom n{n-r}\\
&=\sum_{r=0}^n n\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+(n-r)\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r=0}^n (n-r)\binom n{n-r}^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r'=0}^n r'\binom n{r'}^2
&&(r'=n-r)\\
&=2\sum_{r=0}^n r\binom nr^2\\
\sum_{r=0}^n r\binom nr^2
&=\frac n2\binom {2n}n\color{red}\blacksquare\end{align}$$
A: $1.$ Counting Argument:


*

*For the RHS:


*

*$\tbinom {2n}n$ counts the ways to do some thing.


*For the LHS:


*

*Observe that $\tbinom n k ^2 = \tbinom n k\tbinom n{n-k}$.

*$\tbinom n k\tbinom n{n-k}$ counts the way to do some other thing.

*$\sum_{k=0}^n \tbinom n k\tbinom n{n-k}$ counts the ways to do this other thing for all $k\in\{0,\ldots,n\}$ .


*Compare. If you are doing the same thing, the count of ways must be the same.


And such...
