Difference between revisions of "2005 AMC 10A Problems/Problem 9"

(Solution)
(See Also)
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Therefore the desired [[probability]] is <math>\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}</math>
 
Therefore the desired [[probability]] is <math>\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}</math>
  
==See Also==
+
==See also==
*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=8|num-a=10}}
  
*[[2005 AMC 10A Problems/Problem 8|Previous Problem]]
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[[Category:Introductory Number Theory Problems]]
 
 
*[[2005 AMC 10A Problems/Problem 10|Next Problem]]
 
 
 
*[[Combination]]
 
[[Category:Introductory Combinatorics Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:28, 13 August 2019

Problem

Three tiles are marked $X$ and two other tiles are marked $O$. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads $XOXOX$?

$\mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3}$

Solution

There are $\frac{5!}{2!3!}=10$ distinct arrangements of three $X$'s and two $O$'s.

There is only $1$ distinct arrangement that reads $XOXOX$

Therefore the desired probability is $\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}$

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 10 Problems and Solutions

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