A relation on a set A is said to be symmetric iff (a,b)∈A⇒(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 A×A 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