Difference between revisions of "2008 AMC 12B Problems/Problem 22"
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So the probability she can park is <cmath>1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.</cmath> | So the probability she can park is <cmath>1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.</cmath> | ||
| − | (Bijection: When elements of two sets are perfectly paired with each other, i.e each and every element from both sets has exactly one match in the other, and no elements are left out. In the context of this problem, this means the number of distinct ways to order the cars such that no two spaces are adjacent is exactly the number of ways to pick 4 spots out of 13.) | + | (Bijection: When elements of two sets are perfectly paired with each other, i.e. each and every element from both sets has exactly one match in the other, and no elements are left out. In the context of this problem, this means the number of distinct ways to order the cars such that no two spaces are adjacent is exactly the number of ways to pick 4 spots out of 13.) |
==See Also== | ==See Also== | ||
Revision as of 09:36, 25 July 2020
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?
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 ![\[1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.\]](http://latex.artofproblemsolving.com/5/8/d/58d8d8218d347f2f680c6e9126a441c5d42abdfb.png)
(Bijection: When elements of two sets are perfectly paired with each other, i.e. each and every element from both sets has exactly one match in the other, and no elements are left out. In the context of this problem, this means the number of distinct ways to order the cars such that no two spaces are adjacent is exactly the number of ways to pick 4 spots out of 13.)
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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 12 Problems and Solutions | |
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