Difference between revisions of "2007 AMC 8 Problems/Problem 4"

(Solution)
(Solution)
Line 11: Line 11:
 
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 \cdot 5</math> ways he can enter the house by one window and leave
+
So, Georgie has a total of 6*5 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.
+
Therefore, we have D:30 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}}

Revision as of 18:22, 1 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?

$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36$

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 6*5 ways he can enter the house by one window and leave by a different window.

Therefore, we have D:30 ways

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png