A permutation counts the number of ordered subsets of a finite number of objects. Not to be confused with a combination, which counts the number or unordered subsets of a finite number of objects.
The number of ordered subsets of size $m$ from a set of $n \geq m$ distinct objects is $n!(n-m)!$.
Examples
Example 1
If I have a bag containing three balls, one blue, one red, and one yellow, there are $3!/(3 - 3)! = 3! = 3 \cdot 2 \cdot 1 = 6$ distinct orders in which I can randomly draw them. Intuitively, you can think of this as: there are $3$ options when I draw the first ball, $2$ options when I draw the second, and only $1$ option when I draw the third, so we can apply the $mn$ Rule to see there are $3 \cdot 2 \cdot 1 = 6$ distinct orderings.
Example 2
Using the last example, there are $3!/(3 - 2)! = 3!/1! = 3 \cdot 2 \cdot 1 = 6$ distinct orders in which I can randomly draw just two of the three balls. It seems weird since this is the same number of orderings as the last example, but there are, again, $3$ options for the first choice and $2$ options for the second choice. So, by the $mn$-rule, there are $3 \cdot 2 = 6$ distinct orderings.