Answer to Question #142446 in Combinatorics | Number Theory for Peace College

Question

36 students are members of a sports club. Every two of them are either friends or enemies.
(Friendship and enmity are reciprocal, i.e. if A is a friend to B then B is a friend to A, and the
same applies to being enemies.) It has turned out that each of the students has exactly 8 enemies. Let us called a group of three students concurrent if they are either pairwise enemies or pairwise friends to each other. What is the maximum possible quantity of concurrent student
triples in this sports club? (Two distinct concurrent student triples may have mutual students in
them.)

Answer 1

The group where the students are pairwise enemies includes 9 students.

How many concurrent student triples can we choice from this group?

$$\dbinom{9}{3}=\dfrac{9!}{3!(9-3)!}=\dfrac{9(8)(7)}{1(2)(3)}=84$$

The group where the students are friends includes 27 students.

How many concurrent student triples can we choice from this group?

$$\dbinom{27}{3}=\dfrac{27!}{3!(27-3)!}=\dfrac{27(26)(25)}{1(2)(3)}=2925$$

What is the maximum possible quantity of concurrent student

triples in this sports club? 

$$84+2925=3009$$