AMC10 2013 A

AMC10 2013 A · Q7

AMC10 2013 A · Q7. It mainly tests Basic counting (rules of product/sum), Combinations.

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?

一名学生必须从英语、代数、几何、历史、艺术和拉丁语课程中选择四个课程组成课程计划。该计划必须包含英语，并且至少有一门数学课程。有多少种方式可以选择这个课程计划？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 16 16

Answer

Correct choice: (C)

正确答案：（C）

Solution

Because English is required, the student must choose 3 of the remaining 5 courses. If the student takes both math courses, there are 3 possible choices for the final course. If the student chooses only one of the 2 possible math courses, then the student must omit one of the 3 remaining courses, for a total of $2 \cdot 3 = 6$ programs. Hence there are $3 + 6 = 9$ programs.

因为英语是必修的，学生必须从剩下的5门课中选择3门。如果学生选了两门数学课，最后一门课有3种选择。如果学生只选一门数学课（2种选择），则必须从剩下的3门非数学课中省略一门，总共$2 \cdot 3 = 6$种方案。因此总共有$3 + 6 = 9$种方案。

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