TITLE: How do I show that $\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$?
QUESTION [7 upvotes]: $$\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$$
I know how to "prove" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?

REPLY [5 votes]: A combinatorial proof
How many ways can we pick $n$ objects from a selection of $2n$?
Consider the $2n$ objects to be $$(1,1), (1,2), (2,1), (2,2), \dots, (n, 1), (n, 2)$$
Then the number of ways we can pick out $n$ of these is equal to the number of ways we can pick out $n$ conditioning on how many we select with second coordinate $1$.
That is, $$\sum_{k=0}^n \text{ways to select $k$ with $1$-coordinate} \times \text{ways to select $n-k$ with $2$-coordinate}$$
The summand is clearly $$\binom{n}{k} \times \binom{n}{n-k}$$
But that is $$\binom{n}{k} \times \binom{n}{k} = \binom{n}{k}^2$$