Question

Tough (maybe?) question from a math competition: 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?

Answer 1

Just to start things off... They have to have the same number of friends. They can't have 0 friends each because "some are friends with each other." They can't have 5 friends because "not all are friend with each other"
Ways they can each have 1:
There are 6 people and I need to create 3 pairs... nCr(6,2) * nCr(4,2)* nCr(2,2)

Answer 2

Probably should have mentioned, the choices were (A) 60 (B) 170 (C) 290 (D) 320 (E) 660

Answer 3

We will use casework.
Case 1: Each person has 1 friend:
All we have to do is count the number of ways to pair up the people. This can be done in
Case 2: Each person has 2 friends.
Think about this problem geometrically, with each person resembling a point,and a friendship resembling a line segment. One case is that we have two mutually exclusive groups of 3 mutual friends. This looks like two triangles. We can form such in ways. Another case is that we have people arranged in a hexagon in some way. We can form a hexagon in ways.

The case for three or four friends is the same as the case for 2, or 1 friends respectively, because we can just reverse each persons friendships and "nonfriendships" to form 2 or 1 friends for each person.

Thus we have

Sol. by Dynamite127 AOPS

Answer 4

a)

\[{(6!)\over(4!)}*2=60\]

Answer 5

I'm probably wrong, but hey, I worked hard on that answer. I long to be corrected.