From the table, we can see that room no. 301 is preferred by only 1 manager. Hence, company will allocate room no. 301 to M5.

From the table, we can see that room no. 302 is preferred by 3 managers. Hence, company will allocate room no. 302 to anyone of three managers namely M1, M2 and M4. Remaining two managers won't able to get their best preference.

From the table, we can see that room no. 302 is preferred by only 1 manager. Hence, company will allocate room no. 303 to M3. 

Hence, we can say that at most 3 managers will be able to get their best preference. Therefore, option C is the correct answer. 

From the table, we can see that room no. 302 is preferred by 3 managers. Hence, company will allocate room no. 302 to anyone of three managers namely M1, M2 and M4. Remaining two managers won't able to get their best preference. The remaining two manager will get their second preference as shown in the table below, 

Hence, the minimum sum = 1 + 2 + 1 + 2 + 1 = 7. Therefore, option C is the correct answer. 

From the table, we can see that 2 managers have preferred room no 301 and same is true for room no. 302. Hence, only one out of the two manager will get his/her preferred choice as room no. 301. The same is true for room no 302. Room no. 303 will be be straightaway allocated to M3. 302 can allocated to M4 and 301 can be allocated to M6. Then 304 and 305 can be allocated to M1 and M5 respectively. 

Hence, the minimum sum of the preference ranking allotted to all five managers = 1+1+1+3+3 = 9. Hence, option E is the correct answer. 

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

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

Satya Sadhan even as other members have 5 friends.