Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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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 \choose 3 }}</math><math>=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}</math> | 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 \choose 3 }}</math><math>=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}</math> | ||
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| + | ==Solution 3== | ||
| + | |||
| + | Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is <math>\frac{21 - 6\21) = </math>\frac{5/7}. | ||
==See Also== | ==See Also== | ||
Revision as of 00:44, 12 November 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?
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Contents
Solution 1
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
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 
Solution 3
Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is $\frac{21 - 6\21) =$ (Error compiling LaTeX. Unknown error_msg)\frac{5/7}.
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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