Question

一名学生必须从英语、代数、几何、历史、美术和拉丁语中选择四门课。课程组合必须包含英语，并且至少包含一门数学课。有多少种选择方式？

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

$6$

$8$

$9$

$12$

$16$

答案：C

知识点：组合补集计数

难度评级：1140

解答：

英语必须选，所以还要从代数、几何、历史、美术、拉丁语这 $5$ 门中选 $3$ 门。

这样的选法共有 $\binom53=10$ 种，但其中只有历史、美术、拉丁语这一种不含数学课。

因此有效选法为 $10-1=9$ 种。

所以正确答案是 C。

English is required, so choose the other $3$ courses from the $5$ courses Algebra, Geometry, History, Art, and Latin.

There are $\binom53=10$ such choices, but one of them, History-Art-Latin, contains no mathematics course.

Therefore the number of valid programs is $10-1=9$ .

Thus, C is the correct answer.

2013 AMC 10A 第 7 题

答案：C

难度评级：1140

解答：

其他年份的第 7 题