2004 AMC 8 Problems/Problem 4

Problem

Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.

Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

Solution

There are $\binom{4}{3}$ ways to choose three starters. Thus the answer is $\boxed{\textbf{(B)}\ 4}$ .

Solution 2

We can choose $3$ people by eliminating one from a set of $4$ one at a time and the other three get selected. There are $4$ ways to remove a person from a group of four (without considering order), so there are $\textbf{(B)}4$ ways to choose three people, where order doesn't matter.

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 8 Problems/Problem 4

Contents

Problem

Solution

Solution 2

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