Difference between revisions of "2020 AMC 12B Problems/Problem 15"

(Problem)
Line 4: Line 4:
  
 
<math>\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\  13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15</math>
 
<math>\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\  13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15</math>
 +
 +
==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 <math>\boxed{C}</math>.
 +
 +
~FANYUCHEN20020715
  
 
==See Also==
 
==See Also==

Revision as of 01:01, 8 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

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
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