$\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}?$

Question

How do I show that for $n \geq 0$,

$$\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}?$$

I know that $\sum_{k=0}^n \binom{n}{k} = 2^n$, but does this really tell me anything? Thanks.

Answer 1

To choose $n$ elements from a set $S$ of size $2n$, we may partition $S$ into subsets $A$ and $B$ each of size $n$, and then choose $k$ elements from $A$ and $n-k$ elements from $B$, where $0\le k\le n$. Hence, we find that

$$\sum_{k=0}^n \binom{n}{k}\binom{n}{n-k} = \binom{2n}{n}$$

But of course $\binom{n}{k}=\binom{n}{n-k}$.

Answer 2

HINT:

$$(1+x)^{2n}=(1+x)^n(x+1)^n$$

Expand and compare the coefficient of $x^n$

Answer 3

Split $2n$ things into two groups. Pick $k$ from the first group; leave behind $k$ from the second.