AMC 12 Mock Exam 3 Problem 6

Problem 6 of 25

A club has $6$ juniors, two of whom are Alex and Bree, and $5$ seniors. A $4$ -person committee is to be chosen that includes at least one junior and at least one senior. Alex and Bree have a scheduling conflict, so the committee may not include both of them. How many different committees are possible?

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Solution

There are $\binom{11}4=330$ committees in all; $\binom64=15$ are all juniors and $\binom54=5$ are all seniors, leaving $310$ with both classes represented.

Among these, a committee containing both Alex and Bree consists of them plus $2$ of the other $9$ people, and it has a senior unless both extra members are among the other $4$ juniors: $\binom92-\binom42=36-6=30$ such committees. The answer is $310-30=\boxed{280}$ .

Tagged: committees with restrictions (complementary counting)

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