2005 AMC 10A Problems/Problem 9
Contents
Problem
Three tiles are marked and two other tiles are marked
. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads
?
Solution 1
There are distinct arrangements of
s and
s, and only
distinct arrangement that reads
.
Therefore the desired probability is simply .
Solution 2
Arranging the
s and
s in a row is equivalent to fitting the
s into the gaps between the
s.
There are a total of gaps between the
s, so fitting
s into these gaps gives
possible outcomes:
1. The
s are put into different gaps. In this case, the number of possible arrangements is
.
2. The
s are put into the same gap. In this case, as there are
gaps, we simply obtain
possible arrangements.
Therefore the probability of an arrangement that reads is
.
~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 |
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