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 $XOXOX$ ?

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

Solution1

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 $\boxed{\textbf{(B) }\frac{1}{10}}$

Solution2

Imagine you need to fit the two Os into the gaps between the three Xs.

The gaps between the Xs are: _X_X_X_, a total of 4.

You need to fit two Os in the gaps. There are two possible outcomes:

1. The two Os are put into different gaps, in this case the number of arrangements is 4 x 3 / 2 = 6

2. The two Os are put into the same gap, in this case there will be an extra 4.

Therefore the probability of arrangements that reads XOXOX is 1 / (4 + 6) = 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

Solution1

Solution2

See also