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Difference between revisions of "2020 AMC 10B Problems/Problem 17"
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==Video Solution==
 
==Video Solution==
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https://youtu.be/3BvJeZU3T-M
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https://youtu.be/3BvJeZU3T-M (for AMC 10)
 +
https://youtu.be/0xgTR3UEqbQ (for AMC 12)
  
 
~IceMatrix
 
~IceMatrix

Revision as of 06:19, 17 February 2020

The following problem is from both the 2020 AMC 10B #17 and 2020 AMC 12B #15, so both problems redirect to this page.

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

Let us use casework on the number of diagonals.

Case 1: $0$ diagonals There are $2$ ways: either $1$ pairs with $2$, $3$ pairs with $4$, and so on or $10$ pairs with $1$, $2$ pairs with $3$, etc.

Case 2: $1$ diagonal There are $5$ possible diagonals to draw (everyone else pairs with the person next to them.

Note that there cannot be 2 diagonals.

Case 3: $3$ 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: $5$ diagonals There is $1$ way to do this.

Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.

Video Solution

https://youtu.be/3BvJeZU3T-M (for AMC 10) https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~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
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

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