An even number will end in 2, 4, 6, 8, or 0. Because there are even numbers that end in 0, so we need to account for any overlap. Recall the inclusion/exclusion principle:
[tex]|\text{numbers we want}|=|\text{even}|+|\text{no 0s}|-|\text{even AND no 0s}|[/tex]
where [tex]|X|[/tex] is used to denote the size/cardinality of the set [tex]X[/tex].
Any 4 digit positive even integer will have 9 choices for its first digit (1-9, can't start with 0), 10 choices for its second and third digits, and 5 choices for its last digit. So
[tex]|\text{even}|=9\cdot10^2\cdot5=4,500[/tex]
Numbers that contain no 0s have 9 choices for each of their digits, so
[tex]|\text{no 0s}|=9^4=6,561[/tex]
Finally, even numbers not containing 0s have 9 choices for the first three digits and 4 for the last digit.
[tex]|\text{even AND no 0s}|=9^3\cdot4=2,916[/tex]
So counting the numbers we want, we find that there are [tex]4,500+6,561-2,916=8,145[/tex] of them.
