Factorials and Combinations is always an intimidating topic for me. Help?
Case 1: If the digits can't appear twice, there are permutation 10 taken 2 = 90 attempts.
Case 2: If the digits can appear twice, there are
Case 1: There are 10 possible digits for the first missing digit (0 to 9) and then on the other missing digit, there will be only 9 choices since no digits must appear twice in the number. By counting principle, there 10 x 9 ways of arranging the last two digits. You can also view it as permutation of 10 taken 2, 10 is the number of possible digits and 2 is the number of digits.
Case 2: There are 10 possible digits for the first missing digit and there are also 10 possible digits for the second missing digit, since there is no restriction. By the counting principle, there are 10x10 ways of arranging the last two digits.