2013 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2013 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10A solutions, or check the answer key.

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Concepts:combinations complementary counting

Difficulty rating: 1140

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?

$6$

$8$

$9$

$12$

$16$

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.

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