Difference between revisions of "2012 AMC 10A Problems/Problem 23"
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Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen? | Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen? | ||
− | <math> \ | + | |
+ | <math> \text{(A)}\ 60\qquad\text{(B)}\ 170\qquad\text{(C)}\ 290\qquad\text{(D)}\ 320\qquad\text{(E)}\ 660 </math> | ||
== Solution == | == Solution == | ||
Note that if <math>n</math> is the number of friends each person has, then <math>n</math> can be any integer from <math>1</math> to <math>4</math>, inclusive. | Note that if <math>n</math> is the number of friends each person has, then <math>n</math> can be any integer from <math>1</math> to <math>4</math>, inclusive. | ||
+ | |||
+ | Also note that one person can only have at most 5 friends. | ||
Also note that the cases of <math>n=1</math> and <math>n=4</math> are the same, since a map showing a solution for <math>n=1</math> can correspond one-to-one with a map of a solution for <math>n=4</math> by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with <math>n=2</math> when compared to configurations of <math>n=3</math>. Thus, we have two cases to examine, <math>n=1</math> and <math>n=2</math>, and we count each of these combinations twice. | Also note that the cases of <math>n=1</math> and <math>n=4</math> are the same, since a map showing a solution for <math>n=1</math> can correspond one-to-one with a map of a solution for <math>n=4</math> by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with <math>n=2</math> when compared to configurations of <math>n=3</math>. Thus, we have two cases to examine, <math>n=1</math> and <math>n=2</math>, and we count each of these combinations twice. | ||
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For <math>n=2</math>, there are two possibilities. The group of <math>6</math> can be split into two groups of <math>3</math>, with each group creating a friendship triangle. The first person has <math>\binom{5}{2} = 10</math> ways to pick two friends from the other five, while the other three are forced together. Thus, there are <math>10</math> triangular configurations. | For <math>n=2</math>, there are two possibilities. The group of <math>6</math> can be split into two groups of <math>3</math>, with each group creating a friendship triangle. The first person has <math>\binom{5}{2} = 10</math> ways to pick two friends from the other five, while the other three are forced together. Thus, there are <math>10</math> triangular configurations. | ||
− | However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon [[ | + | However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon [[without loss of generality]]. This person has <math>\binom{5}{2} = 10</math> choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are <math>10 \cdot 3! = 60</math> hexagonal configurations, and in total <math>70</math> configurations for <math>n=2</math>. |
As stated before, <math>n=3</math> has <math>70</math> configurations, and <math>n=4</math> has <math>15</math> configurations. This gives a total of <math>(70 + 15)\cdot 2 = 170</math> configurations, which is option <math>\boxed{\textbf{(B)}\ 170}</math>. | As stated before, <math>n=3</math> has <math>70</math> configurations, and <math>n=4</math> has <math>15</math> configurations. This gives a total of <math>(70 + 15)\cdot 2 = 170</math> configurations, which is option <math>\boxed{\textbf{(B)}\ 170}</math>. | ||
+ | |||
+ | === Note === | ||
+ | |||
+ | We can also calculate the triangular configurations by applying <math>\frac{\binom{6}{3}}{2} = \frac{20}{2}=10</math> (Because choosing <math>A</math>,<math>B</math> and <math>C</math> is the same as choosing <math>D</math>,<math>E</math> and <math>F</math>. | ||
+ | |||
+ | |||
+ | For the hexagonal configurations, we know that the total amount of combinations is <math>6!=720</math>. However, we must correct for our overcounting because of rotation and reflection. We have that there are <math>{720 \over 6 \cdot 2} = 60</math> because there <math>6</math> rotations of the hexagon and 2 reflections (clockwise and counterclockwise) for each valid way. | ||
+ | |||
+ | === Note 2 === | ||
+ | We can also calculate the hexagonal configurations by placing each person at a random vertex (6!) and dividing for rotations (by 6) and reflections (by 2): | ||
+ | <math>\frac{6!}{12}=60</math> | ||
+ | |||
+ | |||
+ | |||
+ | Lamsheeper | ||
+ | |||
+ | === Another Way === | ||
+ | |||
+ | For the hexagonal configuration, it is essentially congruent to counting the number of paths that start and end with A. From <math>A</math>, we have <math>5</math> options, then the next one has <math>4</math>, and so on, so we have: <math>5\cdot 4 \hdots 1 \cdot 1</math>, where the final <math>1</math> is because the final point needs to reconnect to <math>A</math>. Thus, there are <math>5! = 120</math> ways. Dividng by <math>2</math> since we overcounted by accounting for both directions, we have <math>120/2 = \fbox{60}</math> | ||
+ | |||
+ | -bigbrain123 | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://www.youtube.com/watch?v=92X9ePsgPRU | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2012|ab=A|num-b=22|num-a=24}} | ||
+ | |||
{{AMC12 box|year=2012|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2012|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:34, 10 August 2021
- The following problem is from both the 2012 AMC 12A #19 and 2012 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?
Solution
Note that if is the number of friends each person has, then can be any integer from to , inclusive.
Also note that one person can only have at most 5 friends.
Also note that the cases of and are the same, since a map showing a solution for can correspond one-to-one with a map of a solution for by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with when compared to configurations of . Thus, we have two cases to examine, and , and we count each of these combinations twice.
For , if everyone has exactly one friend, that means there must be pairs of friends, with no other interconnections. The first person has choices for a friend. There are people left. The next person has choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are configurations with .
For , there are two possibilities. The group of can be split into two groups of , with each group creating a friendship triangle. The first person has ways to pick two friends from the other five, while the other three are forced together. Thus, there are triangular configurations.
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality. This person has choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are hexagonal configurations, and in total configurations for .
As stated before, has configurations, and has configurations. This gives a total of configurations, which is option .
Note
We can also calculate the triangular configurations by applying (Because choosing , and is the same as choosing , and .
For the hexagonal configurations, we know that the total amount of combinations is . However, we must correct for our overcounting because of rotation and reflection. We have that there are because there rotations of the hexagon and 2 reflections (clockwise and counterclockwise) for each valid way.
Note 2
We can also calculate the hexagonal configurations by placing each person at a random vertex (6!) and dividing for rotations (by 6) and reflections (by 2):
Lamsheeper
Another Way
For the hexagonal configuration, it is essentially congruent to counting the number of paths that start and end with A. From , we have options, then the next one has , and so on, so we have: , where the final is because the final point needs to reconnect to . Thus, there are ways. Dividng by since we overcounted by accounting for both directions, we have
-bigbrain123
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=92X9ePsgPRU
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.