Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 AMC 10A Problems/Problem 9: Difference between revisions

← Older edit
mNo edit summary
Sevenoptimus (talk | contribs)
editor
698 edits
Improved formatting and explanations
 
(15 intermediate revisions by 9 users not shown)
Line 2: Line 2:
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?


<math> \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} </math>
<math>
\textbf{(A) } \frac{1}{12}\qquad \textbf{(B) } \frac{1}{10}\qquad \textbf{(C) } \frac{1}{6}\qquad \textbf{(D) } \frac{1}{4}\qquad \textbf{(E) } \frac{1}{3}
</math>


==Solution==
==Solution 1==
There are <math>\frac{5!}{2!3!}=10</math> distinct arrangments of three <math>X</math>'s and two <math>O</math>'s.  
There are <math>\frac{5!}{2!3!} = 10</math> distinct arrangements of <math>3</math> <math>X</math>s and <math>2</math> <math>O</math>s, and only <math>1</math> distinct arrangement that reads <math>XOXOX</math>.


There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math>
Therefore the desired probability is simply <math>\boxed{\textbf{(B) }\frac{1}{10}}</math>.


Therfore the desired [[probability]] is <math>\frac{1}{10} \Rightarrow \mathrm{(B)}</math>
==Solution 2==
Arranging the <math>3</math> <math>X</math>s and <math>2</math> <math>O</math>s in a row is equivalent to fitting the <math>O</math>s into the gaps between the <math>X</math>s.


==See Also==
There are a total of <math>3+1 = 4</math> gaps between the <math>X</math>s, so fitting <math>2</math> <math>O</math>s into these gaps gives <math>2</math> possible outcomes:
*[[2005 AMC 10A Problems]]


*[[2005 AMC 10A Problems/Problem 8|Previous Problem]]
1. The <math>2</math> <math>O</math>s are put into different gaps. In this case, the number of possible arrangements is <math>\binom{4}{2} = 6</math>.


*[[2005 AMC 10A Problems/Problem 10|Next Problem]]
2. The <math>2</math> <math>O</math>s are put into the same gap. In this case, as there are <math>4</math> gaps, we simply obtain <math>4</math> possible arrangements.


*[[Combination]]
Therefore the probability of an arrangement that reads <math>XOXOX</math> is <math>\frac{1}{4 + 6} = \boxed{\textbf{(B) }\frac{1}{10}}</math>.
[[Category:Introductory Combinatorics Problems]]
 
~Dew grass meadow
 
==See also==
{{AMC10 box|year=2005|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}

Latest revision as of 16:14, 1 July 2025

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 $XOXOX$?

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

Solution 1

There are $\frac{5!}{2!3!} = 10$ distinct arrangements of $3$ $X$s and $2$ $O$s, and only $1$ distinct arrangement that reads $XOXOX$.

Therefore the desired probability is simply $\boxed{\textbf{(B) }\frac{1}{10}}$.

Solution 2

Arranging the $3$ $X$s and $2$ $O$s in a row is equivalent to fitting the $O$s into the gaps between the $X$s.

There are a total of $3+1 = 4$ gaps between the $X$s, so fitting $2$ $O$s into these gaps gives $2$ possible outcomes:

1. The $2$ $O$s are put into different gaps. In this case, the number of possible arrangements is $\binom{4}{2} = 6$.

2. The $2$ $O$s are put into the same gap. In this case, as there are $4$ gaps, we simply obtain $4$ possible arrangements.

Therefore the probability of an arrangement that reads $XOXOX$ is $\frac{1}{4 + 6} = \boxed{\textbf{(B) }\frac{1}{10}}$.

~Dew grass meadow

See also

2005 AMC 10A (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_9&oldid=253058"