I am reading a combinatorial proof for the identity ${2n\choose n}={n\choose 0}^2+{n\choose 1}^2+⋯+{n\choose n}^2$. The proof I read resorts to the concept of bit string.
Personally I don't like the introduction of bit string since it only increases complexity. My way to prove the identity is as: ${2n\choose n}$ implies choose $n$ objects from $2n$ objects. This is equivalent to partitioning $2n$ objects into $2$ equal groups, and, in the first case we choose $0$ from one group and $n$ from another; In the second case we choose $1$ from one group and $n−1$ from another; in the third case we choose $2$ from one group and $n−2$ from another. Continue this fashion, we have ${2n\choose n}={n\choose 0}{n\choose n}+{n\choose 1}{n\choose n-1}+{n\choose 2}{n\choose n−2}+⋯+{n\choose n}{n\choose 0}$.
Since ${n\choose k}={n\choose n−k}$, the formula we get above can be rewritten as ${2n\choose n}={n\choose 0}^2+{n\choose 1}^2+{n\choose 2}^2+⋯+{n\choose n}^2$.
Is my idea correct?