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Is it true that $\forall n,k \in \mathbb{N}$, the number of k-element subsets of an n-element set is \begin{gather*} \binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2) ... (n-k+1)}{k!} \end{gather*}
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