number of cases:
(X is any digit except 6)
6XX = 9X9 = 81
X6X = 8x9 = 72 , can't have 0 at the front
XX6 = 8x9 = 72
66X = 9
6X6 = 9
X66 = 8
Number of cases = add them up
how many whole numbers from 100 to 1000 contain the digit 6 exactly twice?
Sorry about the other post, it's wrong.
1 answer