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2008 AMC 12B Problem 22

Problem 22 of 25HarderCombinatorics

A parking lot has 1616 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 22 adjacent spaces. What is the probability that she is able to park?

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Solution

After the 1212 cars park, 44 spaces are empty, equally likely to be any 44 of the 16,16, for (164)=1820\binom{16}{4} = 1820 equally likely sets. Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place 44 non-adjacent empties among 1616 is (134)=715.\binom{13}{4} = 715. So the probability she can park is 1−7151820=11051820=1728. 1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}. Thus, the correct answer is E.
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Tagged: complementary counting · combinations · arrangements with restrictions

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