@Puja Mishra. Suppose 4 committees are A,B,C and D .
Now suppose there is a member of the club, say Y. Now he has to choose 2 committees out of 4.
In how many ways he can choose? ^{4}C_{2} = 6 ways.
1)AB 2)AC 3)AD 4)BC 5)BC 6)CD
Y can choose any 1 of these combinations. Suppose he chooses AB. So AB has a member in common- Y. Both the conditions are satisfied.
Now suppose there is another member of the club, say Z. Z can't choose combination AB.Choosing AB will violate condition (ii). In this way, Each [possible combinations of the committee has a different member in common.