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2013 AMC 10A Problem 7

Problem 7 of 25EasierCombinatorics

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?

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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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Tagged: combinations · complementary counting

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