> [!NOTE] > Let $n$ be a natural number. Then $\sum_{k=0}^{n} \binom{n}{k}^{2} = \binom{2n}{n} $where $\binom{n}{k}$ denotes a [[Binomial Coefficient|binomial coefficient]]. ###### Combinatorial proof (Partitioning by number of elements in $[n]$)