2019 elements?

The number of ways we can assign a different element of F(A) to each number of A, is the same as the number of ways we can put 2019 distinct elements into order: 2019!

Since we can choose any of the 2019 elements of the range to any of the 2019 elements in the range, total number of A to A functions is: $2019^{2019}$

Hence, the required proportion:

$\frac {2019!}{ 2019^{2019} } = \frac { \color{#BBBBBB} \require { cancel } \cancel {\color{#3D99F6} 2019 \times } \color{#333333} 2018! } { \color{#BBBBBB} \cancel { \color{#3D99F6} 2019 \times } \color{#333333} 2019^{2018} } = \boxed { \frac {2018!}{2019^{2018} } }$

Let's look at all the possible functions first. We know we have 2019 elements in $A$ , meaning if we want to build a function from A to A, we have for each element a total of 2019 diffrent images, hence $2019^{2019}$ . But we want to know how much of these functions are one-to-one. So again, for the first element in $A$ , we have a toal of 2019 elements to map. Though, this time we have for $a_2$ , only 2018 because if we would map it the same image as of $a_1$ , our function wouldn't be one-to-one. Hence we would get that the amount of total one-to-one functions is $2018!$ , writing it as fraction and reducing we get: $\frac{2019!}{2019^{2019}}=\frac{2018!}{2019^{2018}}$