Difference between revisions of "2020 AMC 10B Problems/Problem 17"
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Thus, in total there are <math>2+5+5+1=\boxed{\textbf{(C) }13}</math> possible ways. | Thus, in total there are <math>2+5+5+1=\boxed{\textbf{(C) }13}</math> possible ways. | ||
− | + | ~Pearl2008 (Minor Edits) | |
==Solution 2== | ==Solution 2== | ||
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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. | 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. | ||
− | (This is invalid. Imagine a circle of numbers. <math>1</math>,<math>2</math>,<math>3</math>,<math>4</math>,<math>5</math>,<math>6</math>,<math>7</math>,<math>8</math>,<math>9</math>,<math>10</math>. You might think that <math>1</math>,<math>10</math>,<math>2</math>,and <math>6</math> is such a "<math>4</math>-person group", but <math>6</math> does not know <math>10</math> or <math>2</math>. Despite the answer being correct, this solution does not work. (Arcticturn or anybody, please correct me if I am wrong. I only THINK this solution is invalid and have provided my reasoning.)) | + | (This is invalid. Imagine a circle of numbers. <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, <math>10</math>. You might think that <math>1</math>, <math>10</math>, <math>2</math>,and <math>6</math> is such a "<math>4</math>-person group", but <math>6</math> does not know <math>10</math> or <math>2</math>. Despite the answer being correct, this solution does not work. (Arcticturn or anybody, please correct me if I am wrong. I only THINK this solution is invalid and have provided my reasoning.)) |
~Arcticturn | ~Arcticturn | ||
+ | |||
-PerseverePlayer (Things in parenthesis) | -PerseverePlayer (Things in parenthesis) | ||
+ | |||
+ | (multiple AI's say that PerseverePlayer is correct) | ||
==Video Solution (HOW TO THINK CRITICALLY!!!)== | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
Line 49: | Line 52: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
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==Video Solution by TheBeautyOfMath== | ==Video Solution by TheBeautyOfMath== |
Latest revision as of 10:26, 15 August 2024
- The following problem is from both the 2020 AMC 10B #17 and 2020 AMC 12B #15, so both problems redirect to this page.
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 1
Consider the 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: diameters
There are ways: either
pairs with
,
pairs with
, and so on or
pairs with
,
pairs with
, etc.
Case 2: diameter
There are possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be 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: diameters
There are possible sets of
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
possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the
possibilities by
to get
.
Note that there cannot be a case with diameters because then there would have to be
diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: diameters
There is only way to do this.
Thus, in total there are possible ways.
~Pearl2008 (Minor Edits)
Solution 2
If each person knows exactly people, that means we form "
-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
. The
nd pair is just
. We need to multiply these together since these are
group. The
rd pair would be
. The
th pair is
. We multiply these
together and get
. The final group would be
. So we add these up and we have
possible ways.
(This is invalid. Imagine a circle of numbers. ,
,
,
,
,
,
,
,
,
. You might think that
,
,
,and
is such a "
-person group", but
does not know
or
. Despite the answer being correct, this solution does not work. (Arcticturn or anybody, please correct me if I am wrong. I only THINK this solution is invalid and have provided my reasoning.))
~Arcticturn
-PerseverePlayer (Things in parenthesis)
(multiple AI's say that PerseverePlayer is correct)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution by TheBeautyOfMath
https://youtu.be/3BvJeZU3T-M?t=419
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.