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2020 AMC 10B Problems/Problem 17: Difference between revisions

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<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 1==
==Solution==
Consider the <math>10</math> people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Consider the <imath>10</imath> people to be standing in a circle, where two people opposite of each other form a diameter of the circle.


Let us use casework on the number of pairs that form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.


Case 1: <math>0</math> diameters
Case 1: <imath>0</imath> diameters


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.
There are <imath>2</imath> ways: either <imath>1</imath> pairs with <imath>2</imath>, <imath>3</imath> pairs with <imath>4</imath>, and so on or <imath>10</imath> pairs with <imath>1</imath>, <imath>2</imath> pairs with <imath>3</imath>, etc. So <imath>\boxed{\textbf{2}}</imath> cases.


Case 2: <math>1</math> diameter
Case 2: <imath>1</imath> diameter


There are <math>5</math> possible diameters to draw (everyone else pairs with the person next to them).
There are <imath>5</imath> possible diameters to draw (everyone else pairs with the person next to them). So <imath>\boxed{\textbf{5}}</imath> cases.


Note that there cannot be <math>2</math> diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Note that there cannot be <imath>2</imath> diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.


Case 3: <math>3</math> diameters
Case 3: <imath>3</imath> diameters


There are <math>5</math> possible sets of <math>3</math> diameters to draw.
There are <imath>5</imath> possible sets of <imath>3</imath> diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have <math>5*2=10</math> possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the <math>10</math> possibilities by <math>2</math> to get <math>5</math>.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have <imath>5 \cdot 2=10</imath> possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the <imath>10</imath> possibilities by <imath>2</imath> to get <imath>\boxed{\textbf{5}}</imath>.


Note that there cannot be a case with <math>4</math> diameters because then there would have to be <math>5</math> diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Note that there cannot be a case with <imath>4</imath> diameters because then there would have to be <imath>5</imath> diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.


Case 4: <math>5</math> diameters
Case 4: <imath>5</imath> diameters


There is only <math>1</math> way to do this.
There is only <imath>\boxed{\textbf{1}}</imath> way to do this.


Thus, in total there are <math>2+5+5+1=\boxed{\textbf{(C) }13}</math> possible ways.
Thus, in total there are <imath>2+5+5+1=\boxed{\textbf{(C) }13}</imath> possible ways.
- Minor edits by Pearl2008


==Solution 2==


If each person knows exactly <math>3</math> people, that means we form "<math>4</math>-person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has <math>\dbinom{4}{2}=6</math>. The <math>2</math>nd pair is just <math>\dbinom{2}{2} =1</math>. We need to multiply these together since these are <math>1</math> group. The <math>3</math>rd pair would be <math>\dbinom{4}{2}=6</math>. The <math>4</math>th pair is <math>\dbinom{2}{2}=1</math>. We multiply these <math>2</math> together and get <math>6</math>. The final group would be <math>\dbinom{2}{2}=1</math>. So we add these up and we have <math>6 + 6 + 1 = \boxed{\textbf{(C) }13}</math> possible ways.
~Pearl2008 + Alzwang (Minor Edits)
 
~Arcticturn


==Video Solution (HOW TO THINK CRITICALLY!!!)==
==Video Solution (HOW TO THINK CRITICALLY!!!)==
Line 46: Line 42:


~Education, the Study of Everything
~Education, the Study of Everything


==Video Solution by TheBeautyOfMath==
==Video Solution by TheBeautyOfMath==
Line 55: Line 48:
https://youtu.be/0xgTR3UEqbQ
https://youtu.be/0xgTR3UEqbQ


== Video Solution ==
== Video Solution by Sohil Rathi ==
https://youtu.be/0W3VmFp55cM?t=2796
https://youtu.be/0W3VmFp55cM?t=2796


Latest revision as of 12:01, 11 November 2025

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

Consider the $10$ people to be standing in a circle, where two people opposite of each other form a diameter of the circle.

Let us use casework on the number of pairs that form a diameter of the circle.

Case 1: $0$ diameters

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. So $\boxed{\textbf{2}}$ cases.

Case 2: $1$ diameter

There are $5$ possible diameters to draw (everyone else pairs with the person next to them). So $\boxed{\textbf{5}}$ cases.

Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.

Case 3: $3$ diameters

There are $5$ possible sets of $3$ diameters to draw. Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5 \cdot 2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $\boxed{\textbf{5}}$.

Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.

Case 4: $5$ diameters

There is only $\boxed{\textbf{1}}$ way to do this.

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


~Pearl2008 + Alzwang (Minor Edits)

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/t1WQTZ8TF7k

~Education, the Study of Everything

Video Solution by TheBeautyOfMath

https://youtu.be/3BvJeZU3T-M?t=419

https://youtu.be/0xgTR3UEqbQ

Video Solution by Sohil Rathi

https://youtu.be/0W3VmFp55cM?t=2796

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

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