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

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== Solution 2 ==
 
== 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>\textbf{(B)}4</math> ways to choose three people, where order doesn't matter.
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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

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 $\boxed{\textbf{(B)}\ 4}$ ways to choose three people, where order doesn't matter.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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