Difference between revisions of "2004 AMC 8 Problems/Problem 4"

Revision as of 03:29, 24 December 2012

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?

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

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

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

@@ Line 4: / Line 4: @@
 Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
-<math> \textbf{(A)}2\qquad\textbf{(B)}4\qquad\textbf{(C)}6\qquad\textbf{(D)}8\qquad\textbf{(E)}10 </math>
+<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10 </math>
 == Solution ==
-There are <math>\binom{5}{3}</math> ways to choose three starters. Thus the answer is <math>\boxed{\textbf{(B)}\ 4}</math>
+There are <math>\binom{4}{3}</math> ways to choose three starters. Thus the answer is <math>\boxed{\textbf{(B)}\ 4}</math>.
+==See Also==
+{{AMC8 box|year=2004|num-b=3|num-a=5}}