Counting natural numbers less than 100 containing a 3

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Homework Statement



How many (natural) numbers less than 100 contain a 3? (Note: 13, 35 and 73 all
contain a 3 but 42, 65 and 88 do not).

The Attempt at a Solution



Of course I know that the numbers containing a 3 including 10 numbers starting with a 3 (30, . . . 39),and 10 numbers ending in a 3 (3, 13, . . . , 93), with 33 being counted twice, so a total of 19 numbers. I've found this by counting. But is there a quick systematic way of obtaining this answer using combinations/permutations etc? Unfortunently my knowledge of combinatorics is very poor, so I appreciate any help.

Between 10 to 100 there are 98 2-digit numbers that can possibly contain a 3 in the 1's or 10's positions... I'm stuck here.
 

Discussion

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Not every combinatorics problem involves permutations or combinations.

Think of it this way: You want to count the numbers of the form xy where x or y is a 3.

It would be easy to overcount, so be careful.
 


awkward said:
Not every combinatorics problem involves permutations or combinations.

Think of it this way: You want to count the numbers of the form xy where x or y is a 3.

It would be easy to overcount, so be careful.

The only count would be 33. So other than counting, there are no easy systematic ways of doing this?
 


How many numbers are there where x is a 3?

How many numbers are there where y is a 3?