Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$

I was hoping to find a more “mathematical” proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$.

I already know the logical Proof:

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

Hence summation can be expressed as:

$$\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n}\binom{n}{0}$$

One can think of it as choosing $n$ people from a group of $2n$
(imagine dividing a group of $2n$ into $2$ groups of $n$ people each. I can get $k$ people from group $1$ and another $n-k$ people from group $2$. We do this from $k = 0$ to $n$.


The combinatorial explanation is straightforward. There’s also a roundabout approach through what are called “generating functions.” The binomial theorem tells us that


The $x^n$ coefficient of the above occurs with $c=n$, wherein the coefficient is


However, the $x^n$ coefficient of $(1+x)^n(x+1)^n=(1+x)^{2n}$, again by the binomial theorem, is

$$\binom{2n}{n}. $$

Equating the two gives the result.

Source : Link , Question Author : Lance C , Answer Author : anon

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