Problem: A positive integer is called ascending if, in its decimal representation, there are at least two digits, and each digit is less than any digit to its right. How many ascending positive integers are there? Answer: Note that an ascending number can never have the digit 0. So, we simply choose $k$ digits from $9$ where $k = 2, \ldots, 9$ and for each selection, there is one correct ordering that leaves the corresponding number ascending. Our answer is thus $\binom{9}{9} + \binom{9}{8} + \ldots + \binom{9}{2} = 2^9 - \binom{9}{1} - \binom{9}{0} = 502.$