Difference between revisions of "2020 AMC 12B Problems/Problem 15"
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We have 2 cases. Assume these people are A,B,...,J. | We have 2 cases. Assume these people are A,B,...,J. | ||
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Case 1: Group with people nearby: we have (AB)(CD)(EF)(GH)(IJ) and (AJ)(IH)(GF)(ED)(CB), totally 2 cases. | Case 1: Group with people nearby: we have (AB)(CD)(EF)(GH)(IJ) and (AJ)(IH)(GF)(ED)(CB), totally 2 cases. | ||
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Case 2: (AF) is a group. So here we have 3 sub-cases: | Case 2: (AF) is a group. So here we have 3 sub-cases: | ||
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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 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). | ||
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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 2: (BG)(CH) are groups, the rest groups are (DE)(IJ). We also have 5 cases for the same reason of Sub 1. | ||
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Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case. | Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case. | ||
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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>. | ||
Revision as of 01:01, 8 February 2020
Problem
There are 
people standing equally spaced around a circle. Each person knows exactly 
of the other 
people: the 
people standing next to her or him,as well as the person directly across the circle. How many ways are there for the people to split up into 
pairs so that the members of each pair know each other?
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 
.
~FANYUCHEN20020715
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. 
