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Difference between revisions of "2003 AMC 8 Problems/Problem 16"
 
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==Solution==
 
==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{\textbf{(D)}\ 12}</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{\textbf{(D)}\ 12}</math>.
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 +
==Solution 2 (Quick)==
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If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is <math>\boxed{\textbf{(D)}\ 12}</math>.
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 +
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]
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 +
==Video Solution==
 +
https://youtu.be/GeAxPKwsL5E Soo, DRMS, NM
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=15|num-a=17}}
 
{{AMC8 box|year=2003|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:51, 1 July 2023

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 $\boxed{\textbf{(D)}\ 12}$.

Solution 2 (Quick)

If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is $\boxed{\textbf{(D)}\ 12}$.

Solution by ILoveMath31415926535

Video Solution

https://youtu.be/GeAxPKwsL5E Soo, DRMS, NM

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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