Exercise 1605Combinatorics › Counting principlesL1Added 3 Nov 2026

A combinatorial identity

#combinatorial identity#double counting#binomial

Statement

Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$ .

Hint

Use Vandermonde's identity with $m=n$ and $k=n$ .

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Vandermonde. $\binom{m+n}{k}=\sum_j\binom{m}{j}\binom{n}{k-j}$ .

Apply. With $m=n$ and $k=n$ : $\binom{2n}{n}=\sum_j\binom{n}{j}\binom{n}{n-j}=\sum_j\binom{n}{j}^2$ .

Conclusion. The identity holds.

Combinatorial interpretation. Choose $n$ from $2n$ by taking $j$ from the first half and $n-j$ from the second.