2006 AMC 12B Problems/Problem 7: Difference between revisions
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<math>3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}</math> | <math>3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}</math> | ||
Alternative solution: | |||
If there was no restriction, there would be 4!=24 ways to sit. However, only 2/4 of the people can sit in the driver's seat, so our answer is <math>\frac{2}{4}\cdot 24= 12 \Rightarrow \text{(B)}</math> | |||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}} | ||
Revision as of 11:25, 1 June 2009
Problem
Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?
Solution
First, we seat the children.
The first child can be seated in 
spaces.
The second child can be seated in 
spaces.
Now there are 
ways to seat the adults.
Alternative solution:
If there was no restriction, there would be 4!=24 ways to sit. However, only 2/4 of the people can sit in the driver's seat, so our answer is 
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 AMC 12 Problems and Solutions | |
