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 12B Problems/Problem 15"

(→‎Solution)
(→‎Solution)
Line 18: Line 18:
 
Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case.
 
Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case.
  
−
Thus, the total number of cases are 2+5+5+1=13 cases. So put <math>\boxed{C}</math>.
+
Thus, the total number of cases are 2+5+5+1=13 cases. So put <math>\boxed{C}.</math>
 +
 
  
 
~FANYUCHEN20020715
 
~FANYUCHEN20020715
 +
 +
==Solution 2==
 +
Take cases on the number of pairs of people who shake hand across the table.
 +
Case 1: No pairs. There are two ways.
 +
Case 2: One pair. There's five ways to choose this pair, and afterwards the remaining 8 people are determined -> 5.
 +
Case 3: Two pairs. There are no ways to do this.
 +
Case 4: Three pairs. We see that the three pairs must be adjacent, so there are 5 ways to do this.
 +
Case 5: Four pairs. Again no ways.
 +
Case 6: All five pairs. One way.
 +
 +
Thus, the total number is 2+5+5+1=13 ways, and the anser is <math>\boxed{C}.</math>
 +
 +
~ rzlng
  
 
==See Also==
 
==See Also==

Revision as of 17:48, 9 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

We have 2 cases. Assume these people are A,B,...,J.

Case 1: Group with people nearby: we have (AB)(CD)(EF)(GH)(IJ) and (AJ)(IH)(GF)(ED)(CB), totally 2 cases.

Case 2: (AF) is a group. So here we have 3 sub-cases:

Sub 1: The rest of them group with people nearby, that is (BC)(DE)(GH)(IJ), we have 5 cases since (AF) can be (BG), (CH), (DI) and (EJ).

Sub 2: (BG)(CH) are groups, the rest groups are (DE)(IJ). We also have 5 cases for the same reason of Sub 1.

Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case.

Thus, the total number of cases are 2+5+5+1=13 cases. So put $\boxed{C}.$


~FANYUCHEN20020715

Solution 2

Take cases on the number of pairs of people who shake hand across the table. Case 1: No pairs. There are two ways. Case 2: One pair. There's five ways to choose this pair, and afterwards the remaining 8 people are determined -> 5. Case 3: Two pairs. There are no ways to do this. Case 4: Three pairs. We see that the three pairs must be adjacent, so there are 5 ways to do this. Case 5: Four pairs. Again no ways. Case 6: All five pairs. One way.

Thus, the total number is 2+5+5+1=13 ways, and the anser is $\boxed{C}.$

~ rzlng

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 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_12B_Problems/Problem_15&oldid=117578"