2013 AMC 10A Problem 7

Attempt Problem 7 of the 2013 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

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?
66
88
99
1212
1616
Answer: C
Concepts:combinationscomplementary counting
Small hint:
English is fixed, so choose three of the five remaining courses
Big hint:
Count all three-course choices from the remaining courses, then remove the choice with no math
Solution:
English is required, so choose the other 33 courses from the 55 courses Algebra, Geometry, History, Art, and Latin.
There are (53)=10\binom53=10 such choices, but one of them, History-Art-Latin, contains no mathematics course.
Therefore the number of valid programs is 10−1=910-1=9.
Thus, C is the correct answer.
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