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

Revision as of 09:45, 4 July 2013

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

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.

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 == See also ==
 {{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}}
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Difference between revisions of "2006 AMC 12B Problems/Problem 7"

Revision as of 09:45, 4 July 2013

Problem

Solution

See also