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Difference between revisions of "2003 AMC 8 Problems/Problem 16"

(Created page with "There are only <math>2</math> people who can go in the driver's seat--Bonnie and Carlo. Any of the <math>3</math> remaining people can go in the front passenger seat. There are <...")
 
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==Problem==
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Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has <math>4</math> seats: <math>1</math> Driver seat, <math>1</math> front passenger seat, and <math>2</math> back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24 </math>
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==Solution==
 
There are only <math>2</math> people who can go in the driver's seat--Bonnie and Carlo. Any of the <math>3</math> remaining people can go in the front passenger seat. There are <math>2</math> people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are <math>2\cdot3\cdot2</math> or <math>12</math> ways. The answer is then <math>\boxed{D}</math>.
 
There are only <math>2</math> people who can go in the driver's seat--Bonnie and Carlo. Any of the <math>3</math> remaining people can go in the front passenger seat. There are <math>2</math> people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are <math>2\cdot3\cdot2</math> or <math>12</math> ways. The answer is then <math>\boxed{D}</math>.
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{{AMC8 box|year=2003|num-b=15|num-a=17}}

Revision as of 09:13, 26 December 2011

Problem

Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has $4$ seats: $1$ Driver seat, $1$ front passenger seat, and $2$ back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24$

Solution

There are only $2$ people who can go in the driver's seat--Bonnie and Carlo. Any of the $3$ remaining people can go in the front passenger seat. There are $2$ people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2\cdot3\cdot2$ or $12$ ways. The answer is then $\boxed{D}$.

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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