Difference between revisions of "2003 AMC 8 Problems/Problem 16"
(→Solution) |
|||
| Line 5: | Line 5: | ||
==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>. | ||
| + | |||
| + | ==Solution2== | ||
| + | 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>. | ||
==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}} | ||
Revision as of 19:59, 5 January 2023
Contents
Problem
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 
seats: 
Driver seat, front passenger seat, and 
back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
Solution
There are only people who can go in the driver's seat--Bonnie and Carlo. Any of the 
remaining people can go in the front passenger seat. There are people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are 
or 
ways. The answer is then 
.
Solution2
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 .
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. 
