How many positive, three-digit integers contain at least one $3$ as a digit but do not contain a $5$ as a digit?

One such number is 333.

The digits to be considered are 0,1,2,4,6,7,8,9 as well as 3.

Numbers containing 2 3's : - 3.3 - there are 8 of these; 33. - there are 8 and. 33 - there are 7 (because the first number can't be 0)

Numbers beginning with a 3 + 2 other digits: 3 ... There are 8P2 + 8 (these last 8 are the double digits)

Numbers. 3. There are 7P2 + 7 and Numbers ... 3 there are 7P2 + 7 also.

7P2 = 7! / 5! = 42, 8P2 = 8! / 6! 56

So the answer is 1 + 8 + 8 + 7 + 56 + 8 + 42 + 7 + 42 + 7

= 186