Question

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?

$6$

$8$

$9$

$12$

$16$

Answer: C

Concepts:combinations complementary counting

Difficulty rating: 1140

Solution:

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 Problem 7

Answer: C

Difficulty rating: 1140

Solution:

Problem 7 in Other Years