Difference between revisions of "2007 AMC 8 Problems/Problem 4"
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Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows. | Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows. | ||
| − | So, Georgie has a total of 6 | + | So, Georgie has a total of <math>6 \cdot 5</math> ways he can enter the house by one window and leave |
by a different window. | by a different window. | ||
| − | Therefore, we have D | + | Therefore, we have <math> \boxed{\textbf{(D)}\ 30} </math> ways. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=3|num-a=5}} | {{AMC8 box|year=2007|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:22, 16 August 2021
Problem
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?
Solution
Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.
So, Georgie has a total of 
ways he can enter the house by one window and leave
by a different window.
Therefore, we have 
ways.
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
