Word Problem Guidebook, Chap. 10 Hard Question Set, last problem (believe its #20). Problem asks to calculate total amount of 3 digit numbers with the following conditions: 1) all 3 digits cannot be the same integer, 2) 0 cannot be a digit.
ManhattanGRE solution is 9x9x9 minus 9 (all digits from 111 to 999 that have all 3 digits the same integer) = 720. This makes sense to me however I don’t understand why 9x9x8 is not a solution. In other words, 9 (digits 1-9) x 9 (digits 1-9) x 8 (less one digit so that 3rd digit does match 2 other digits). For example, a menu has 2 appetizers, 2 mains, 2 deserts. The total possible combinations is 2x2x2 = 8. And if you just are allowed one dessert then it is 2x2x1 = 5. I’m not sure why the same logic doesn’t apply to digits question above. Thank you.