2005 AMC 10A Problems/Problem 9

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

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

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2005 AMC 10A Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also