Answer to Question #146605 in Combinatorics | Number Theory for ankit

Question

37 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 pair wise 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

$\text{Total number of groups of 3 students:}$

$\frac{37*36*35}{3*2*1}=7770$

$\text{In groups that do not meet the condition of the task, there will be 1 or 2 enemies}$

$\text{Number of groups with one enemy:}$

$8*\frac{29*28}{2*1}=3248$

$\text{Number of groups with two enemy:}$

$\frac{8*7}{2*1}*29=1044$

$\text{The number of groups corresponding to the condition of the problem:}$

$7770-3248-1044 =3478$

Answer: 3478 concurrent student triples