Entries in a particular lottery game are made up of 3 digits, each 0 through 9. If the order of digits in the entries matters, how many different possible enteries exist in which all 3 digits are not equal? A. 516 B. 720 C. 989 D. 990 E. 1321

Hope I dint miss something. First digit 10 choices including 0 since the problem says 0 can be the first digit 2nd digit->9 choices 3rd digit-> 8 choices 10*9*8 = 720

I did miss something. 2 digits can be equal but not all 3. 10*9*2 =180 But ech of the above picks can be arranged in 3 ways So 180*3 but we need to divide by 2 since some pairs of arrangement are the same I pick 1 2 1 and then 2 1 1 3 different numbers can be formed by either of these 2 picks but the resulting unique numbers is just half. Out of 540 only 270 are unique.(hope I dint mess up here, Ian or some one else please correct me) 180*3/2 = 270 So 720+270 = 990 Nice prob. Regards, CR

alternatively we could solve this by saying total choices - number of choices with all 3 digits same. Total choices = 10*10*10 Choices with all 3 digits same = 10 (incld 000, 111, 222, 333) 1000 - 10 = 990 Voila

alternatively we could solve this by saying total choices - number of choices with all 3 digits same. Total choices = 10*10*10 Choices with all 3 digits same = 10 (incld 000, 111, 222, 333) 1000 - 10 = 990 Well done....

m&m wrote:alternatively we could solve this by saying total choices - number of choices with all 3 digits same. Total choices = 10*10*10 Choices with all 3 digits same = 10 (incld 000, 111, 222, 333) 1000 - 10 = 990 Voila Cool approach !!!