2012 AMC 12A Problem 19

Attempt Problem 19 of the 2012 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

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?

6060

170170

290290

320320

660660

Answer: B
Concepts:graph theorycomplementary countingcasework
Difficulty rating: 2090
Solution:

Model people as vertices of a graph, with edges for friendships. Everyone has the same degree nn with 1n4.1 \le n \le 4. The cases nn and 61n6 - 1 - n are complementary graphs, so n=1n = 1 pairs with n=4n = 4 and n=2n = 2 with n=3.n = 3.

For n=1n = 1 the graph is a perfect matching: 53=155 \cdot 3 = 15 ways. Thus n=4n = 4 also gives 15.15.

For n=2n = 2 the graph is a union of cycles: either two triangles ((52)=10)\left(\binom{5}{2} = 10\right) or one hexagon (6!12=60),\left(\dfrac{6!}{12} = 60\right), totaling 70.70. Thus n=3n = 3 also gives 70.70.

The total is 15+15+70+70=170.15 + 15 + 70 + 70 = 170.

Thus, the correct answer is B.

← Problem 18#18
Full Exam

Problem 19 in Other Years