Difference between revisions of "2020 AMC 12B Problems/Problem 15"
(→Solution) |
(→Solution 2) |
||
| Line 25: | Line 25: | ||
==Solution 2== | ==Solution 2== | ||
Take cases on the number of pairs of people who shake hand across the table. | Take cases on the number of pairs of people who shake hand across the table. | ||
| + | |||
Case 1: No pairs. There are two ways. | 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 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 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 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 5: Four pairs. Again no ways. | ||
| + | |||
Case 6: All five pairs. One way. | Case 6: All five pairs. One way. | ||
Revision as of 17:48, 9 February 2020
Contents
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
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
~ 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. 
