How many integers with at least one eight and at least one nine as digits are th

Treat numbers from 0 to 9999 as 4-digit strings (including leading zeros). Total strings = 10^4 = 10000. Let A be strings with no '8', |A|=9^4=6561. Let B be strings with no '9', |B|=9^4=6561. Intersection (no 8 and no 9): 8^4=4096. By inclusion-exclusion, strings with at least one 8 and at least one 9 = 10000 - (6561+6561-4096) = 974. Since 0 (0000) does not have both, this count applies to 1-9999 as well, and 10000 has no 8 or 9. Thus answer is 974.