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Difference between revisions of "2020 AMC 10B Problems/Problem 17"

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If each person knows exactly 3 people, that means we form "4 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}</math> =6. The 2nd pair is just <math>\dbinom{2}{2}</math> =1. We need to multiply these together since these are 1 group. The 3rd pair would be <math>\dbinom{4}{2}</math> =6. The 4th pair is <math>\dbinom{2}{2}</math> =1. We multiply these 2 together and get 6. The final group would be <math>\dbinom{2}{2}</math> =1. So we add these up and we have <math>6 + 6 + 1 = \boxed{(C)13}</math> possible ways.
 
If each person knows exactly 3 people, that means we form "4 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}</math> =6. The 2nd pair is just <math>\dbinom{2}{2}</math> =1. We need to multiply these together since these are 1 group. The 3rd pair would be <math>\dbinom{4}{2}</math> =6. The 4th pair is <math>\dbinom{2}{2}</math> =1. We multiply these 2 together and get 6. The final group would be <math>\dbinom{2}{2}</math> =1. So we add these up and we have <math>6 + 6 + 1 = \boxed{(C)13}</math> possible ways.
 +
 +
~Arcticturn
  
 
==See Also==
 
==See Also==

Revision as of 18:35, 27 March 2021

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 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.

Case 2: $1$ diameter

There are $5$ possible diameters to draw (everyone else pairs with the person next to them).

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 diameters to draw.

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 $1$ way to do this.

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

Video Solution

https://youtu.be/3BvJeZU3T-M?t=419 (for AMC 10) https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~IceMatrix

Video Solution

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

~ pi_is_3.14

Solution 4

If each person knows exactly 3 people, that means we form "4 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 $\dbinom{4}{2}$ =6. The 2nd pair is just $\dbinom{2}{2}$ =1. We need to multiply these together since these are 1 group. The 3rd pair would be $\dbinom{4}{2}$ =6. The 4th pair is $\dbinom{2}{2}$ =1. We multiply these 2 together and get 6. The final group would be $\dbinom{2}{2}$ =1. So we add these up and we have $6 + 6 + 1 = \boxed{(C)13}$ possible ways.

~Arcticturn

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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