Difference between revisions of "2020 AMC 10B Problems/Problem 17"
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==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 21:06, 7 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
Let us use casework on the number of diagonals.
Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.
Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.
Note that there cannot be 2 diagonals.
Case 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: diagonals There is way to do this.
Thus, in total there are possible ways.
Video Solution
~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 |
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