Difference between revisions of "2008 AMC 12B Problems/Problem 22"
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==Solution== | ==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. | 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 <cmath>1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}=\frac{17}{28} | + | 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}=\boxed{\textbf{(E)}\frac{17}{28}}.</cmath> |
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2008|ab=B|num-b=21|num-a=23}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:21, 2 February 2015
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 ![\[1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}=\boxed{\textbf{(E)}\frac{17}{28}}.\]](http://latex.artofproblemsolving.com/7/1/0/710e37e36ca6c735ec8bf14ea9774d0228c8bc79.png)
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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