Central binomial coefficient identity

Question

$$\sum_{k=0}^n \frac{(2n)!}{(k! (n-k)!)^2} = \binom{2n}{n}^2$$

The original combinatorial interpretation was something like how many strings of length $2n$ with 4 letters A,B,C,D have $k$ A's, $k$ B's, $n-k$ C's, and $n-k$ D's. From looking at the first few values, this looks like the square of the number of strings of length $2n$ with 2 letters, $n$ A's and $n$ B's.

Is there a nice combinatorial proof for this? I was thinking something like taking two strings with $n$ A's and $n$ B's, for example AAABBB and ABABAB, then mapping each pair of letters from both strings to one of four new letters. Then there are $k$ AA's, $n-k$ AB's, $k$ BB's, and $n-k$ BA's. Is this correct? Is there a nice algebraic proof too?

Answer 1

Your combinatorial argument seems fine. Here's an algebraic proof: $$\sum_{k=0}^n \frac{(2n)!}{(k! (n-k)!)^2}$$ $$= (2n)!\sum_{k=0}^n \frac{1}{(k! (n-k)!)^2}$$ $$= \frac{(2n)!}{n!n!}\sum_{k=0}^n \frac{(n!)^2}{(k! (n-k)!)^2}$$ $$ = \binom{2n}{n}\sum_{k=0}^n \binom{n}{k}^2$$ It is a well-known fact that $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$. This yields the required expression.