From a group of $7$ persons, $7$ committees are formed. Any $2$ committees have exactly $1$ member in common. Each person is in exactly $3$ committees.
- At least $1$ committee must have more than $3$ members;
- Each committee must have exactly $3$ members;
- Each committee must have more than $3$ members;
- Nothing can be said about the sizes of the committees.
I used block designs. Using which I could prove that it is not possible for a committee to have 4 members .Which lead to me the answer that each committee must have exactly 3 members. However, I was looking to see if there existed a more formal approach for solving this.
My approach was just in case there are four members in the committee then they should each be in two more committees . If none of the two are in the same committee then this makes the number of committees to be 9. That will be in contradiction just seven committees existing . Hence there should be some committees such that two participants of the original four membered committee belong in them . Hence we come at a contradiction again.