How many numbers of 4 digits contains at least one "5"? How many numbers of $4$ digits contains at least one "$5$"?

My suggestion is:

First I count all the numbers of type 5xxx (where x can be a number between 0 and 9). There is 999.

Then I add all the numbers of type y5xx (where y can be 1, 2, 3, 4, 6, 7, 8, 9). There is 8*99

Then I add all the numbers of type yz5x (where z can be 0, 1, 2, 3, 4, 6, 7, 8, 9). There is 8*9*9

Then I add all the numbers of type yzz5. There is 8*9*9

Total: 999+8*99+8*9*9+8*9*9 = 3087

My problem is, I'm not sure if this is correct OR if there exist a smarter method.
HINT: Compute how many 4 digit numbers contain no 5, and subtract that from the total count of 4 digit numbers.
If first digit is 5 the rest can't be five
1*9*9*9

If the second digit is five the rest can't be five and the first can't be zero
8*1*9*9

If the third digit is five the rest can't be five and the first can't be zero
8*9*1*9

If the fourth digit is five the rest can't be five and the first can't be zero
8*9*9*1

So in conclusion,
(1*9*9*9) + (8*1*9*9) + (8*9*1*9) + (8*9*9*1) = 2673