a1. Digits in strictly ascending order: It's 9 digits choose 3. 9C3 = 84. That's because there is only one way to arrange 3 digits in ascending order. It's 9C3 not 10, because 0 cannot be used at all since the first digit must be at least 1, and the other two digits must be greater. Answer: 84 ------------------------------------ a2. Digits in strictly descending order: 0 can be used, but it must come last. Case 1. No 0's That's the same as a1, or 84, because we can rearrange each one in descending order. Case 2. There is a zero last. This is the number of two digit numbers containing no zeros, because we can put a 0 on the right end of each and have a three-digit number in descending order. The number of two digit numbers in descending order is 9 digits choose 2 or 9C2 = 36, because for every choice there is only one way to arrange them in descending order. (Then we put a 0 at the end of each). Answer: 84+36 = 120. ------------------------ b1. Digits in non-decreasing order: Note that we cannot use 0. Case 1: All three digits different. It's the 84 in the first part above. Case 2: Those with exactly two digits alike. We can choose the digit to have two alike in 9 ways We can choose the digit that is not like the other two in 8 ways. That's 9*8 = 72 ways. Case 3: Those with all three digits the same. They are 111,222,...,999 There are 9 of those. Answer: 84+72+9 = 165 ----------------------------------------- b2. Digits in non-increasing order. Case 1: No zeros. That the same as the 165 in b1 above, for we can arrange each one in non-increasing order. Notice that there were no 0's. Case 2. Exactly 1 zero, which must be at the right end. That's the same as the 36 of case 2 of a2, arranging the first two digits in non-increasing order, plus the 9 where the first two can be the same: 110,220,...,990 That's 36+9=45 Case 3. Exactly 2 zeros. That's these 9: 100,200,...900 Answer: 165+45+9 = 219 Edwin