How do you find cardinality of the set of functions $A\to A$? How to take the first step to approach this?

Question

If $A$ is set with $10$ elements and $|A| = 10$:

$1)$ $|\{f \;|\; f: A \to A \text{ is a function}\}|=$

$2)$ $|\{f \;|\; f: A \to A \text{ is a bijective function}\}|=$

Answer 1

HINT.

$1.$ To be a function, you need to know what to do with each element. For instance, if I ask you what $f(1)$ is, you have to be able to tell me. Call these $10$ elements the numbers $1$, $2$, ... , $10$. How many options do you have for $f(1)$? $f(2)$? Etc. How many total functions do you then have?

$2$. You now need the function to be bijective. So you need eventually need to choose each number $1$, ... , $10$ at least once (so that $f$ is surjective). But once you choose a number for $f(\text{something})$, you cannot choose it again (so that $f$ is injective). Let's just do this in order. How many options for $f(1)$? Given your choice for $f(1)$, how many options for $f(2)$? Repeat. How many total bijections then do you have?

Answer 2

Let $A = \{a_1, \dots, a_{10}\}$, and $f: A \to A$ be a bijection. Notice that if we list $(a_1, a_2, \dots, a_{10})$, then since $f$ is a bijection $(f(a_1), f(a_2), \dots, f(a_{10}))$ is just a rearrangement of $(a_1, a_2, \dots, a_{10})$. So you are counting the number of ways you can rearrange $(a_1, a_2, \dots, a_{10})$ (keeping in mind that doing nothing is also a rearrangement).

Answer 3

This is not an answer, just a recommendation of a way of looking at things.

When I was a wee lad, mathematically speaking, people denoted the set of maps from a set $A$ to a set $B$ as $B^A$. Try it on some small sets, especially with $|A|=1$ and $|A|=2$.

As to the bijective functions $A$ to $A$, just write them all down, for $A=\{1,2,3\}$. What have you done? You’ve written down all permutations of $A$. Now do you see why my childhood teachers wrote $A!$ ?

Now for an exercise: find the cardinality of the set of all permutations of the natural numbers $\Bbb N$.