Competition problem — Combinatorics

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Combinatorics Difficulty 5.4 AIME, harder Prove it Brazil

A positive integer $n$ is clowny if the number obtained by reversing its digits is greater than $n$ . For example, $2009$ is clowny because $9002$ is greater than $2009$ ; however, $2010$ is not clowny because $0102 = 102$ is less than $2010$ and $3443$ is not clowny because it is equal to the number obtained by reversing its digits. How many four-digit numbers are clowny?

Solution

Let $(abcd)$ be a four-digit number $a$ , $b$ , $c$ , $d$ being its digits. So a number is clowny if and only if $(dcba) > (abcd)$ . This means that either $d > a$ or $d = a$ and $c > b$ . Notice that we cannot have both $d = a$ and $b = c$ because it would imply $(dcba) = (abcd)$ .

There are $\frac{9 \cdot 8}{2} = 36$ choices in the first case (recall that $d > a > 0$ ) and $9 \cdot \frac{10 \cdot 9}{2} = 405$ choices in the second case ( $9$ choices for $a = d$ and $\frac{10 \cdot 9}{2}$ for $c > b \ge 0$ ). So there are $36 + 405 = 441$ four-digit clowny numbers.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.