Difference between revisions of "2012 AMC 10A Problems/Problem 23"

Revision as of 17:04, 26 January 2015

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?

$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 170\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 320\qquad\textbf{(E)}\ 660$

Solution

Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.

Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$ . Thus, we have two cases to examine, $n=1$ and $n=2$ , and we count each of these combinations twice.

For $n=1$ , if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$ .

For $n=2$ , there are two possibilities. The group of $6$ can be split into two groups of $3$ , with each group creating a friendship triangle. The first person has $\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ 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 $\binom{5}{2} = 10$ 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 $10 \cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$ .

As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\cdot 2 = 170$ configurations, which is option $\boxed{\textbf{(B)}\ 170}$ .

It should be as easy as $1,2,3.$

Bold textiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii' == 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.

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 It should be as easy as <math>1,2,3.</math>
-== See Also ==
+'''''Bold text''iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii'
+''== See Also ==
 {{AMC10 box|year=2012|ab=A|num-b=22|num-a=24}}

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Difference between revisions of "2012 AMC 10A Problems/Problem 23"

Revision as of 17:04, 26 January 2015

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