2012 AMC 10A Problems/Problem 23
- The following problem is from both the 2012 AMC 12A #19 and 2012 AMC 10A #23, so both problems redirect to this page.
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 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
.
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 |
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