Difference between revisions of "2004 AMC 8 Problems/Problem 4"
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== Solution == | == Solution == | ||
There are <math>\binom{4}{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>. | ||
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| + | == Solution 2 == | ||
| + | We can choose <math>3</math> people by eliminating one from a set of <math>4</math> one at a time and the other three get selected. There are <math>4</math> ways to remove a person from a group of four (without considering order), so there are <math>\boxed{\textbf{(B)}\ 4}</math> ways to choose three people, where order doesn't matter. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=3|num-a=5}} | {{AMC8 box|year=2004|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 08:17, 15 December 2022
Contents
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 
.
Solution 2
We can choose 
people by eliminating one from a set of 
one at a time and the other three get selected. There are ways to remove a person from a group of four (without considering order), so there are ways to choose three people, where order doesn't matter.
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. 
