Solution:
A relation on a set A is said to be symmetric iff (a,b)inA⇒(b,a)∈A,∀a,b∈A
Here A={3,4,6,8,9}
Number of order pairs of A×A=5×5=25
Divide 25 order pairs of AtimesA in 3 parts as follows:
Part – A : (3, 3), (4, 4), (6, 6), (8, 8), (9, 9)
Part – B : (3, 4), (3, 6), (3, 8), (3, 9), (4, 6), (4, 8),(4, 9), (6, 8), (6, 9), (8, 9)
Part – C : (4, 3), (6, 3), (8, 3),(9, 3), (6, 4), (8, 4), (9, 4), (8, 6), (9, 6), (9, 8)
In part – A, both components of each order pair are same.
In part – B, both components are different but not two such order pairs are present in which first component of one order pair is the second component of another order pair and vice-versa.
In part–C, only reverse of the order pairs of part –B are present i.e., if (a, b) is present in part – B, then (b, a) will be present in part –C
For example (3, 4) is present in part – B and (4, 3) present in part –C.
Number of order pair in A, B and C are 5, 10 and 10 respectively.
In any symmetric relation on set A, if any order pair of part –B is present then its reverse order pair of part –C will must be also present.
Hence number of symmetric relation on set A is equal to the number of all relations on a set D, which contains all the order pairs of part –A and part– B.
Now n(D)=n(A)+n(B)=5+10=15
Hence number of all relations on set D=(2)15
⇒ Number of symmetric relations on set D=(2)15
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