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>. | ||
+ | |||
+ | ==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 <math>\boxed{\textbf{(D)}\ 12}</math>. | ||
+ | |||
+ | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] | ||
+ | |||
+ | ==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}} |
Latest revision as of 23:51, 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 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
.
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 .
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.