2008 AMC 12B Problems/Problem 22

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Problem 22

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

$\textbf{(A)} \; \frac {11}{20} \qquad \textbf{(B)} \; \frac {4}{7} \qquad \textbf{(C)} \; \frac {81}{140} \qquad \textbf{(D)}...$

Solution

Auntie Em won't be able to park only when none of the four available spots touch. We can form a bijection between all such cases and the number of ways to pick four spots out of 13: since none of the spots touch, remove a spot from between each of the cars. From the other direction, given four spots out of 13, simply add a spot between each. So the probability she can park is $1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}=\frac{17}{2...$

2008 AMC 12B (Problems • Resources)
Preceded by Problem 21	Followed by Problem 23
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2008 AMC 12B Problems/Problem 22

From AoPSWiki

Problem 22

Solution

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