Question 26342
Because 1,2,3 must be in the 6 digits we only need to worry about 3 other digits, and how many possible combinations there for them. There are 9 possible digits and we are taking 3 at a time (remember we are working with combinations, order does not matter) So there are 9C3 combinations for the remaining three digits. When we combine all the possible combinations we get 1x1x1x9C3 to get the total 84 combinations