**Base 10 refers to the numbering system in common use that uses decimal numbers.**

We count the number of 3-digit numbers with (i) at least one 5 and having no 3 and (ii) at

least one 5 and having exactly one 3 separately.

(i) Here we first count the whole set and subtract the number of 3-digit numbers having

no 5 from it. Since 3 is not there and 0 cannot be the first digit, we can fill the first digit

in 8 ways. But we can fill the second and third digits in 9 ways(as 0 can be included).

Thus we get 8*9 *9 such numbers. If no 5 is there, then the number of such numbers is

7*8*8. Thus the number of 3-digit numbers not containing 3 and having at least one 5 is

(8 * 9 * 9) - (7 * 8 * 8) = 8(81 - 56) = 200.

(ii) If 3 is there as a digit, then it can be the first digit or may be the second or third digit.

Consider those numbers in which 3 is the first digit. The number of such numbers having at

least one 5 is (9 * 9) - (8 * 8) = 81 - 64 = 17. The number of 3-digit numbers in which the

second digit is 3 and having at least one 5 is (8*9) - (7*8) = 16. Similarly, the number of

3-digit numbers in which the third digit is 3 and having at least one 5 is (8*9) - (7*8) = 16.

Thus we get 17 + 16 + 16 = 49 such numbers.

**Therefore the number of 3-digit numbers having at most one 3 and at least one 5 is 200+49 =
**

249.