Here's a solution to problem #0007-a

Case#1: The number has *one* digit. Well, we have 1,2,3,4,5,6,7,8,9. --->9 integers.

Case#2: The number has *two* digits. The number is of the form **ab**, For the first position, **a**, we have 9 choices of non-zero digits: namely 1-9. Now for the second position, **b**, we have 10 choices of any digits 0-9 *except* **a**, so we have 9 choices. --->9*9=81 integers.

Case#3: The number has *three* digits. The number is of the form **abc**. For the first position, we have 9 choices of non-zero digits: namely 1-9.

Now for the second position, **b**, we have 10 choices of any digits 0-9 *except ***a**, so we have 9 choices. For the final digit, **c**, we have 10 choices of any digits 0-9 *except *what we chose to **a **and **b**, so we have 8 choices. --->9*9*8=648 integers.

In total we have 9+81+648=738 positive integers which satisfy the given conditions. All results follow from the *multiplicative counting principle*.