Question

How many members are there in the club?

• A. 5
• B. 7
• C. 15
• D. 16
• E. 20
• A. 2
• B. 3
• C. 5
• D. 7
• J. 8

Answer 1

The Correct Option is D

Step 1: Translate the conditions into graph theory. - Each member is a point (vertex). - Friendship = edge between two vertices. - Condition 1: No two friends have a common friend. - Condition 2: Any two non-friends have exactly 2 common friends. This is a highly restrictive structure, suggesting a strongly regular graph.

Step 2: Analyze from perspective of a chosen member X. Let X have exactly $m$ friends: $A_1, A_2, A_3, \dots, A_m$. - Since no two friends of X can be friends with each other, all $A_i$ are independent (no edges among them). - Any $A_i$ must be connected to X and to exactly one other person outside this set.

Step 3: Account for “2 common friends for every non-friend pair.” - Consider X and another vertex P not connected to X. - X and P must share 2 common friends. Those friends can only come from $\{A_1, A_2, \dots, A_m\}$. - Thus, each non-neighbour of X must be connected to exactly 2 of the $A_i$.

Step 4: Structure of the graph. - Besides X and the $m$ $A_i$’s, we form vertices $B_{ij}$, each connected to exactly two $A_i$’s. - Total vertices = $1 + m + \binom{m}{2} = 1 + m + \frac{m(m-1)}{2}$.

Step 5: Check possible $m$ values. - For $m=3$: $1+3+3=7$ (not valid, since condition fails). - For $m=4$: $1+4+6=11$ (fails on common-friend condition). - For $m=5$: $1+5+10=16$ (valid configuration). - For $m=6$: $1+6+15=22$ (ruled out, as Satya Pramod said fewer than 22 members).

Step 6: Confirm consistency with conversation. - Satya Sadhan: Not all are friends → true. - Satyabrata: Friends have no common friend → holds. - Satyajit: Non-friends have exactly 2 common friends → satisfied by the construction. - Satya Pramod: Fewer than 22 → consistent, as we found 16.

Step 7: Final Answer. Hence, the number of members is \[ \boxed{16} \]

Answer 2

Step 1: Recall the structure from Q47. We determined that the club has 16 members in total. The configuration is based on strongly regular graph conditions: 1. No two friends share a common friend. 2. Any two non-friends have exactly two common friends.

Step 2: Consider Satya Sadhan (say vertex X). Let X be one member. Suppose he has $m$ direct friends: $A_1, A_2, \dots, A_m$. - None of the $A_i$’s are connected to each other (since friends cannot share a friend). - Each pair of $A_i$ and $A_j$ defines exactly one vertex $B_{ij}$ who is connected to both.

Step 3: Count the total number of vertices. The structure is: \[ 1 + m + \binom{m}{2} \] where: - 1 = X, - $m$ = X’s direct friends, - $\binom{m}{2}$ = the $B_{ij}$ vertices created from every friend-pair. We know the total = 16. \[ 1 + m + \frac{m(m-1)}{2} = 16 \]

Step 4: Solve for $m$. \[ m + \frac{m(m-1)}{2} = 15 \] \[ \frac{m^2 + m}{2} = 15 \] \[ m^2 + m = 30 \] \[ m^2 + m - 30 = 0 \] \[ m = 5 \quad \text{(valid root)} \]

Step 5: Interpret. Thus, Satya Sadhan (and every other member) has exactly 5 friends. \[ \boxed{5} \]