Question

若一个正整数的十进制表示至少有两位，并且每一位数字都小于其右边的任意一位数字，就称它为递增正整数。共有多少个递增正整数？

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?

答案

答案：502

知识点：子集数字组合

难度评级：1700

小提示：

一旦选定一组非零数字，它们的排列顺序就唯一确定

Once a set of nonzero digits is chosen, their order is forced

大提示：

排除大小为 $0$ 和 $1$ 的子集；这里的全集是 $\{1,2,\ldots,9\}$

Exclude subsets of sizes $0$ and $1$ from the subsets of $\{1,2,\ldots,9\}$

解答：

数字 $0$ 不可能出现，因为它必须位于首位，而十进制表示不含前导零。从 $\{1,\ldots,9\}$ 中任选至少两个数字，并按递增顺序排列，就恰好得到一个递增正整数。因此，所求数目为 $\begin{aligned}2^9-\binom90-\binom91&=512-1-9\\&=502\end{aligned}\text{。}$

The digit $0$ cannot occur, because it would have to be the first digit and leading zeroes are not part of a decimal representation. Every subset of at least two digits from $\{1,\ldots,9\}$ gives exactly one ascending integer when written in increasing order. Hence the number is $\begin{aligned}2^9-\binom90-\binom91&=512-1-9\\&=502.\end{aligned}$

1992 AIME 第 2 题

答案：502

难度评级：1700

小提示：

大提示：

解答：

其他年份的第 2 题