Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10B Problems/Problem 17"

m
(→‎Solution)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
 +
Let us use casework on the number of diagonals.
 +
 +
Case 1: <math>0</math> diagonals
 +
There are <math>2</math> ways: either <math>1</math> pairs with <math>2</math>, <math>3</math> pairs with <math>4</math>, and so on or <math>10</math> pairs with <math>1</math>, <math>2</math> pairs with <math>3</math>, etc.
 +
 +
Case 2: <math>1</math> diagonal
 +
There are <math>5</math> possible diagonals to draw (everyone else pairs with the person next to them.
 +
 +
Note that there cannot be 2 diagonals.
 +
 +
Case 3: <math>3</math> diagonals
 +
 +
Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.
 +
 +
Case 4: <math>5</math> diagonals
 +
There is <math>1</math> way to do this.
 +
 +
Thus, in total there are <math>2+5+5+1=\boxed{13}</math> possible ways.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:06, 7 February 2020

Problem

There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?

$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$

Solution

Let us use casework on the number of diagonals.

Case 1: $0$ diagonals There are $2$ ways: either $1$ pairs with $2$, $3$ pairs with $4$, and so on or $10$ pairs with $1$, $2$ pairs with $3$, etc.

Case 2: $1$ diagonal There are $5$ possible diagonals to draw (everyone else pairs with the person next to them.

Note that there cannot be 2 diagonals.

Case 3: $3$ diagonals

Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.

Case 4: $5$ diagonals There is $1$ way to do this.

Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.

Video Solution

https://youtu.be/3BvJeZU3T-M

~IceMatrix

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_17&oldid=117112"