Difference between revisions of "2004 AMC 8 Problems/Problem 4"
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Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen? | 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 == | == Solution == | ||
| − | There are <math>\binom{ | + | 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}} | ||
| + | {{MAA Notice}} | ||
Revision as of 00:54, 5 July 2013
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?
Solution
There are 
ways to choose three starters. Thus the answer is 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
