Difference between revisions of "2018 AMC 8 Problems/Problem 23"

(I did not edit but I want to ask: Are you sure that is the stars and bars formula? I don't think that is the stars and bars formula)
m (Solution)
Line 15: Line 15:
<math>\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}</math>
<math>\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}</math>
===Solution 1===
We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent.
If all <math>3</math> points are adjacent, then we have <math>8</math> choices. If we have exactly <math>2</math> adjacent points, then we will have <math>8</math> places to put the adjacent points and also <math>4</math> places to put the remaining point, so we have <math>8\cdot4</math> choices. The total amount of choices is <math>{8 \choose 3} = 8\cdot7</math>.
Thus our answer is <math>\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}</math>
===Solution 2 ===
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>
===Solution 3 (Stars and Bars)===
Let <math>1</math> point of the triangle be fixed at the top. Then, there are <math>{7 \choose 2} = 21</math> ways to chose the other 2 points. There must be <math>3</math> spaces in the points and <math>3</math> points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and <math>2</math> extra points (k-1) distributed so by the stars and bars formula, <math>{n+k-1 \choose k-1}</math>, there are <math>{4 \choose 2} = 6</math> ways to arrange the bars and stars. Thus, the probability is <math>\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}</math>.
The stars and bars formula might be inaccurate
==Video Solution==
==Video Solution==

Revision as of 04:44, 9 November 2020


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?

[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]

$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$


Video Solution

https://www.youtube.com/watch?v=VNflxl7VpL0 - Happytwin

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS