Use a combinatorial argument to prove a combinatorial identity

Question

$$\dbinom{2n}{n}=\sum_{k=1}^{n}\dbinom{n}{k}^2$$

Attempt: RHS gives the number of ways to choose some committee and then choose another committee such that some people may be in two of the committees. The LHS shows the scenario when n people are picked from a group of 2n. I am having trouble showing how the LHS shows the same idea as the RHS.

Answer 1

Hint $$\dbinom{n}{k}=\dbinom{n}{n-k}$$

$$\dbinom{2n}{n}=\sum_{k=0}^{n}\dbinom{n}{k}^2$$

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