Answer:
5 different numbers, 2 even
Step-by-step explanation:
Since we have 5 digits and they are non repeating we can use the permutation formula:
[tex]P(n, k) = \frac{n!}{k!}, $where $ k \leq n ; k, n \in \mathbb{N}[/tex]
It gives us the amount of ways to arrange k objects out of n objects without repetition, but order does matter.
In our case the amount of 4 digit numbers that can be formed using the digits from 1 to 5 is: [tex]P(5, 4) = \frac{5!}{4!} = 5.[/tex]
And out of those 5, only 2 are even
