Counting and the probability of functions from A to itself

Question

Suppose a set A with N elements, I have to translate the statement:

The relation R corresponds to an injective function from A to itself

into a logical statement, and also find the probability of it holding.

For the first part, I tried ∀a ∈ A, ∀b ∈ A, (a, b) ∈ R ⇒ ( a = b ), which I am not completely certain on.

For the probability, I believe the number of injective functions would be N!, so then would the probability then be N! / ${N}^{N}$ ?

I also have a follow up question, where instead the statement in question is:

The relation R corresponds to a bijective function from A to itself such that given any input, the output is not the same as the input.

This one I am completely lost with. How should I be going about answering this question?

Answer 1

For the first part, I tried ∀a ∈ A, ∀b ∈ A, (a, b) ∈ R ⇒ ( a = b ), which I am not completely certain on.

No, that says that any element is at most related to itself. (If an element is related to anything that thing is itself).

A relation is an injective function when every element in the domain maps to a single distinct element in the codomain (a one-to-one mapping)

For the probability, I believe the number of injective functions would be N!, so then would the probability then be $N! / N^N$ ?

Indeed. The codomain of an injective function is a permutation of the domain.

The relation R corresponds to a bijective function from A to itself such that given any input, the output is not the same as the input.

This is a derangment.