Question:
Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE, three members each interacted with four members of AXIAL and the remaining members each interacted with two members of AXIAL. The greatest possible number of company BELIANCE members in the party is:
Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE, three members each interacted with four members of AXIAL and the remaining members each interacted with two members of AXIAL. The greatest possible number of company BELIANCE members in the party is:
Show Hint
Count total interactions from both sides of the party; the AXIAL side gives a hard ceiling of 24, and you need the largest n for which the BELIANCE side stays within that ceiling.
Updated On: Jul 10, 2026
- 9
- 10
- 11
- 12
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The Correct Option is A
Solution and Explanation
Step 1: Model the situation as a bipartite graph.
Think of each interaction between an AXIAL member and a BELIANCE member as an edge connecting them. We are given the number of interactions (degree) of certain people on both sides.
Step 2: Count total interactions from the BELIANCE side.
Let n be the total number of BELIANCE members at the party. Of these, 3 members interacted with 4 AXIAL members each, giving \(3 \times 4 = 12\) interactions. The remaining \((n-3)\) members interacted with 2 AXIAL members each, giving \(2(n-3)\) interactions. So the total interactions, counted from the BELIANCE side, is \[ 12 + 2(n-3) = 2n + 6 \]
Step 3: Bound the total interactions from the AXIAL side.
There are 8 AXIAL members, and no AXIAL member interacted with more than 3 BELIANCE members. So the total interactions, counted from the AXIAL side, is at most \(8 \times 3 = 24\).
Step 4: Equate and solve, since both sides count the same interactions.
\[ 2n + 6 \leq 24 \] \[ 2n \leq 18 \] \[ n \leq 9 \]
Step 5: Check that n = 9 is actually achievable.
At n = 9, the total interactions equal exactly 24, so each of the 8 AXIAL members must interact with exactly 3 BELIANCE members, the maximum allowed. This can be arranged by spreading the connections evenly among the 9 BELIANCE members, so n = 9 is a genuine maximum, not just a theoretical bound.
Final Answer:
The greatest possible number of BELIANCE members is \[ \boxed{9} \]
Think of each interaction between an AXIAL member and a BELIANCE member as an edge connecting them. We are given the number of interactions (degree) of certain people on both sides.
Step 2: Count total interactions from the BELIANCE side.
Let n be the total number of BELIANCE members at the party. Of these, 3 members interacted with 4 AXIAL members each, giving \(3 \times 4 = 12\) interactions. The remaining \((n-3)\) members interacted with 2 AXIAL members each, giving \(2(n-3)\) interactions. So the total interactions, counted from the BELIANCE side, is \[ 12 + 2(n-3) = 2n + 6 \]
Step 3: Bound the total interactions from the AXIAL side.
There are 8 AXIAL members, and no AXIAL member interacted with more than 3 BELIANCE members. So the total interactions, counted from the AXIAL side, is at most \(8 \times 3 = 24\).
Step 4: Equate and solve, since both sides count the same interactions.
\[ 2n + 6 \leq 24 \] \[ 2n \leq 18 \] \[ n \leq 9 \]
Step 5: Check that n = 9 is actually achievable.
At n = 9, the total interactions equal exactly 24, so each of the 8 AXIAL members must interact with exactly 3 BELIANCE members, the maximum allowed. This can be arranged by spreading the connections evenly among the 9 BELIANCE members, so n = 9 is a genuine maximum, not just a theoretical bound.
Final Answer:
The greatest possible number of BELIANCE members is \[ \boxed{9} \]
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