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 <math>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 <math> \boxed{\textbf{(D)}\ 30} </math> ways. | |
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/XdnS_5KEx6s | ||
+ | |||
==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}} |
Latest revision as of 16:30, 28 October 2024
Contents
[hide]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.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by WhyMath
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.