Number of permutations

Statement

Let $X$ be a finite set with $|X| = n$. Then the number of permutations of $X$ is $n!$.

Remarks

For $n = 3$ there are $3! = 6$ permutations of $\{1, 2, 3\}$: $123$, $132$, $213$, $231$, $312$ and $321$. The count follows from the rule of product. The image of the first element has $n$ choices, the image of the second has $n - 1$ choices, and so on. It is the case $k = n$ of the number of injections.

Explanations

L1 human

To arrange $n$ different objects in a row, you have $n$ choices for the first place, then $n - 1$ for the second, and so on. The product of these numbers is $n!$.