Order of the symmetric group

Statement

Let $n$ be a positive integer. The symmetric group $S_n$ has order $n!$.

Remarks

A permutation $\sigma$ has $n$ choices for $\sigma(1)$, then $n - 1$ choices for $\sigma(2)$, and so on, down to $1$ choice for $\sigma(n)$. For example, $|S_3| = 6$ and $|S_4| = 24$.

Explanations

L1 human

To make a permutation, you choose where each symbol goes, one symbol after the other. Each choice removes one place from the next choices. So the number of permutations is $n \cdot (n-1) \cdots 1 = n!$.