# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. 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 members of company $BELIANCE$ in the party is:
 ✔ A. $9$ ✖ B. $10$ ✖ C. $11$ ✖ D. $12$

Solution:
Option(A) is correct

The important constraint here is that in the party, no member of $AXIAL$ had interacted with more than three members of $BELIANCE$.

Given that there are 8 members of company $AXIAL$ and three members of company $BELIANCE$ interacted with four members of $AXIAL$.

Therefore, the maximum possible number of members of company $BELIANCE$ in the party will be $3+ {^4C_2}$ (Since, each of the remaining members of the company $BELIANCE$ have interacted with two members of $AXIAL$) $= \textbf{9}$

## (4) Comment(s)

Tyler
()

Given number of members in AXIAL = 8

“In the party no member of AXIAL had interacted with more than three members of BELIANCE” => Maximum number of interactions would be = 8*3 = 24.

“Out of all the members of BELIANCE, three members- each interacted with four members of AXIAL” =>number of interactions of these 3 members of BELIANCE= 3*4 = 12 => remaining number of interactions for BELIANCE = 24-12 = 12

“and the remaining members – each interacted with two members of AXIAL” => remaining number of members in BELIANCE = 12/2 = 6

Maximum number of members in BELIANCE = 6+3 = 9

Ankit
()

"4C2" in solution. Thank you

Ankit
()

Can you please explain it more clearly?

Anita
()

Which part of the question/solution is not clear to you?