The question says relation on A.So cross-product (A*A) to get all possible relations on A.
A={1,2,3}
A*A={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)};n(A)=3C1 * 3C1 = 9
Relation matrix
A*A |
1 |
2 |
3 |
1 |
(1,1) |
(1,2) |
(1,3) |
2 |
(2,1) |
(2,2) |
(2,3) |
3 |
(3,1) |
(3,2) |
(3,3) |
If it's represented in matrix form,where reflexive relations are the principal diagonal entries and symmetric relations are elements which are symmetric about the principal diagonal.
Now,any subset of this crossproduct is a relation on A.Total number of subsets = 2^9=512. But we are restricted by some constraints from the question as follows:
1)Neither it's reflexive(or, all the principal diagonal elements are present in the relation set) nor it's irreflexive(or, all the principal diagonal elements are absent)-So, we can either choose one of the 3 principal diagonal elements or 2 of the principal diagonal elements,which is same as (3C1 + 3C2).
2) Relation must be Symmetric(if aRb is present then bRa must be present in the relation set)-There are 3 symmetric entries about the principal diagonal {(1,2) or (2,1) and (1,3) or (3,1) and (2,3) and (3,2)}.So it becomes a problem of counting how many subsets are possible with 3 symmetric entries(i.e 2^3=8)(Note:- principal diagonal elements are symmetric by default).
Therefore,total no. of ways to form a relation is achieved by combining 1) & 2) i.e. first select set which satisfy 1) and then select set which satisfies 2).(Note:- order of selecting can be switched).
8*(3C1 + 3C2)=8*6=48 ways.