1992 AIME Problems/Problem 2

Revision as of 19:12, 26 July 2006 by 4everwise (talk | contribs) (Solution)

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?

Solution

First, we line up the nine digits that may be used: $\displaystyle 1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $\displaystyle 2^9=512$ working numbers.

Subtracting off the one-digit numbers and the null set, our answer is $\displaystyle 512-10=502.$

See also

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