Instructions

Answer question based on the following information

Every Saturday, the members of Raja Harish Chandra Club meet in the evening. All the member of the club are honest and never lie. Last Saturday, the following conversation was heard at one of the table with five members sitting around it.

Satya Sadhan: In this club not all members are friends with each other.
Satyabrata: None of the pair of friends in this club has any common friend.
Satyajit: Every pair of member who are not friends has exactly two common friends in this club.
Satya Pramod: There are fewer than 22 people in this club.

Question 47

How many members are there in the club?

Solution

We can represent People by points and relation of being friends by joining those points . so a set of people and their relations can be represented by a graph .
1. No two friends have a common friend .
2.Any two non friends have exactly 2 common friends .

Consider any point X 
let say there are exactly m points $$\ A_1,\ A_2,A_3.....A_m\ \ $$ connected with X .None of these points can be connected with each other , therefore corresponding to every pair Ai , Aj there have to be exactly two points connected to both.
One is X, the other is Bij.This has to be different values for i,j . This has to be the complete graph i.e a point x , a point connected to X or a point connected to Ai .
There can be no point unconnected with every one of these points . If X and P are unconnected they have to have two common friends .
All friends of X are included in $$\ A_1,\ A_2,A_3.....A_m\ \ $$
So P has to be connected with some of A's . Also there can be no other point connected with any Ai .If there is a point B it can't be connected to any Aj for it would be then Bij and X and B will have 1 common friend
so the points in graphs are :
$$X,\ A_1,\ A_2,A_3.....A_m\ B_{12\ },\ B_{13}......B_{\left(m-1\right)m}\ \ $$
The number of points is
n = 1+m +(m)(m-1)/2
some values of m, n are given below :


we have to consider m =3,4,5
When m =3
we can't provide friend to B12 so m cannot be 3
m =4
we get

Take B12 , he has two friends A1 and A2 , but we cannot provide two more friends among Bij's . Only B34 is available .

Consider m =5 

B12 has A1 and A2 and we have to provide three more they are (B34 , B35 and B45)
Hence m =5 and n =16
so we get 16 members in the group


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