Ace Test Series: Set Theory & Algebra - Relations

Question

Let $A=\left \{ 1,2,3 \right \}$. Number of relation on $A$ which are neither reflexive, nor irreflexive but symmetric is ___________

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Ans given 48

but I got 8

Please verify

Comments

48 is correct answer...

neither reflexive nor irreflexive ===> all reflexive pairs can't present at a time. And all reflexive pairs can't absent at a time.

there are three reflexive pairs, so 3c0 and 3c3 avoid ==> (3c1+3c2)

for symmetric, mutual pairs are (9-3)/2 = 3.

total = 2^3 . (3c1+3c2)

Answer 1

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.