There are 9C3 ways to choose the 3 digits. Of those 3 choose the digit to be used only twice 3C1 ways. There are 8 places in an 8-digit number for the digits to go. Choose the 2 places for the digit to be used twice in 8C2 ways. That leaves 6 places for the larger of the 2 remaining digits. Choose the 3 places for the larger remaining digit to go in 6C3 ways. That leaves 3 places for the smaller remaining digit. Choose the places for the smallest digits to go in 3C3 or 1 way. (Only 1 way, because there are only 3 places left for the smallest digit). Answer: (9C3)(3C1)(8C2)(6C3)(3C3) = 141120 ways Edwin