Question

Fred's Bank 的 ATM 密码由 $0$ 到 $9$ 中的四个数字组成，允许数字重复。如果密码不能以序列 $9,1,1$ 开头，那么有多少个可能密码？

An ATM password at Fred's Bank is composed of four digits from $0$ to $9,$ with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

$30$

$7290$

$9000$

$9990$

$9999$

答案：D

知识点：乘法原理补集计数

难度评级：1020

视频讲解：

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文字解答：

没有限制时，密码总数为 $10^4$ 。这个条件排除了 $10$ 个密码，因为前 $3$ 位固定为九、一、一，最后一位可以任意。因此可接受密码数为 $10,000 - 10 = 9990.$

所以正确答案是 D。

The total number of passwords with no conditions is $10^4.$ The condition removes $10$ possible passwords since the first $3$ are determined, and the last one can be anything. Therefore, the number of acceptable passwords is $10,000 - 10 = 9990.$

Thus, D is the correct answer.

2016 AMC 8 第 17 题

答案：D

难度评级：1020

视频讲解：

文字解答：

其他年份的第 17 题