Combinations/permutations help

[roam]

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.

 

[awkward]

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.

 

[roam]

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?

 

[Office_Shredder]

How many numbers are there where x is a 3?

How many numbers are there where y is a 3?