Difference between revisions of "1984 AIME Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | First | + | First notice that there is no difference between the maple trees and the oak trees; we have only two types, birth trees and "non-birch" trees. |
| − | The five birch trees must | + | The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making <math>{8\choose5} = 56</math> different ways to arrange this. |
| − | There are | + | There are <math>{12 \choose 5} = 792</math> total ways to arrange the twelve trees, so the probability is <math>\frac{56}{792} = \frac{7}{99}</math>. |
| + | |||
| + | The answer is <math>7 + 99 = \boxed{106}</math>. | ||
== See also == | == See also == | ||
Revision as of 22:16, 5 April 2012
Problem
A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let 
in lowest terms be the probability that no two birch trees are next to one another. Find 
.
Solution
First notice that there is no difference between the maple trees and the oak trees; we have only two types, birth trees and "non-birch" trees.
The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making 
different ways to arrange this.
There are 
total ways to arrange the twelve trees, so the probability is 
.
The answer is 
.
See also
| 1984 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
