Question

how many 4-digit numbers contain at least one 8 or 9

Answer 1

well first of all count all 4-digit numbers, then count all 4-digit numbers with no threes, then subtract the second number from the first.
Right, first the 4-digit numbers. Well, first of all there are 9 possible digits for the first place (you cant use 0, it will become 3 digit number), 10 for the second place (you can use anything from 0 to 9) , 10 for the third place and 10 for the fourth similarly .That makes 9 x 10 x 10 x 10 = 9000.
Now for the 4-digit numbers with no threes. Their first (thousands) digit can be chosen in just 8 ways (no 0 and no 3), their second(hundreds) digit in just 9 (no 3 remember), their third(tens) digit in 9 and their fourth digit in 9 ways similarly. So there are 8 x 9 x 9 x 9 of these. Thats 5832 altogether