Problem:
A parking lot has 16 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 2 adjacent spaces. What is the probability that she is able to park?
Answer Choices:
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Solution:
The four vacant spaces can be located in any of combinations of positions. The arrangements in which Auntie Em is unable to park may be divided into two cases. If the rightmost space is occupied, then every vacant space is immediately to the left of an occupied space. Let denote the union of a vacant space and the occupied space immediately to its right, and let Y denote a single occupied space not immediately to the right of a vacant space. The arrangement of cars and spaces can be represented by a sequence of four X's and eight Y's in some order, and there are possible orders. If the rightmost space is vacant, the arrangement in the remaining 15 spaces can be represented by a sequence of three X's and nine Y's in some order, and there are possible orders. Therefore there are arrangements in which Auntie Em can park, and the requested probability is .
\section*{OR}
Let denote an occupied space, and let denote a vacant space. The problem is equivalent to finding the probability that in a string of 's and 's, there are at least two consecutive 's. Then is the probability that no two 's are consecutive. In a string of 's, there are 13 spaces in which to insert 's to create a string in which no two 's are consecutive. Thus
The problems on this page are the property of the MAA's American Mathematics Competitions