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

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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