Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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==Solution 2 (Complementary)== | ==Solution 2 (Complementary)== | ||
− | We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8 | + | We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8\cdot6-8}{8\cdot7}=\boxed{\textbf{(D) } \frac 57}</math> |
==See Also== | ==See Also== |
Revision as of 14:53, 15 July 2019
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
Solution
We will use constructive counting to solve this. There are cases: Either all
points are adjacent, or exactly
points are adjacent.
If all
points are adjacent, then we have
choices. If we have exactly
adjacent points, then we will have
places to put the adjacent points and also
places to put the remaining point, so we have
choices. The total amount of choices is
.
Thus our answer is
Solution 2 (Complementary)
We can decide adjacent points with
choices. The remaining point will have
choices. However, we have counted the case with
adjacent points twice, so we need to subtract this case once. The case with the
adjacent points has
arrangements, so our answer is
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AJHSME/AMC 8 Problems and Solutions |
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