Question 1128074
You are using the digits 1 through 9 (no zero),
so there are {{{9}}} digits to choose from.
An even number must end in 0, 2, 4, 6, or 8.
Your even three-digit numbers must end in 2, 4, 6, or 8.
That gives you {{{4}}} choices for the last digit.
For each choice, you have {{{9-1=8}}} unused digits to use as second (tens) digit,
to get {{{4*8=32}}} two-digit ending sequences.
Each of those two-digit ending sequence choices
leaves you {{{9-1-1=7}}} unused digits you can choose as the first (hundreds) digit,
for a total of {{{32*7=highlight(224)}}}
three-digit even numbers with no repeated digits.