Difference between revisions of "2006 AMC 12B Problems/Problem 7"
(→See also) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 6: | Line 6: | ||
</math> | </math> | ||
| − | == Solution == | + | == Solution 1 == |
First, we seat the children. | First, we seat the children. | ||
| Line 17: | Line 17: | ||
<math>3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}</math> | <math>3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}</math> | ||
| + | == Solution 2 == | ||
| + | |||
| + | There are only two possible occupants for the driver's seat. After the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus, there are <math>2\cdot 3\cdot 2 = | ||
| + | \boxed{12}</math> possible seating arrangements. ~ aopsav (Credit to AoPS Alcumus) | ||
Alternative solution: | Alternative solution: | ||
| Line 24: | Line 28: | ||
== 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}} | ||
| + | [[Category:Introductory Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:33, 9 April 2024
Contents
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 1
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.
Solution 2
There are only two possible occupants for the driver's seat. After the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus, there are 
possible seating arrangements. ~ aopsav (Credit to AoPS Alcumus)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
