The sum of the squares of the elements of the $n$-th layer of Pascal's triangle is known to be
$$\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$$
Here is the link from wiki.
The proof that I know for this is using the special case of Vandermonde's identity choosing $m=r=n$ below
$$\sum_{k=0}^r \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$$
I was curious if there was a generalization of this to Pascal's simplex (e.g. in 3D. $\sum_{a+b+c=n} \binom{n}{a,b,c}^2 = ?$ where the summands are the squares of the trinomial coefficients). I don't believe there is a way to obtain it from the generalized Vandermonde's identity.