Difference between revisions of "2006 AMC 12B Problems/Problem 7"

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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>
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Alternative solution:
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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>
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== 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 12:25, 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?

$\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48$

Solution

First, we seat the children.

The first child can be seated in $3$ spaces.

The second child can be seated in $2$ spaces.

Now there are $2 \times 1$ ways to seat the adults.

$3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}$


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 $\frac{2}{4}\cdot 24= 12 \Rightarrow \text{(B)}$

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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