If A= x ∈ z^+: x < 10 . and x is a multiple of 3 or 4, where z^+is the set of positive integers, then the total number of symmetric relations on A is

A relation on a set A is said to be symmetric iff $(a, b) \in A \Rightarrow(b, a) \in A, \forall a, b \in A$(a,b)∈A⇒(b,a)∈A,∀a,b∈A Here $A=\{3,4,6,8,9\}$A={3,4,6,8,9} Number of order pairs of $A \times A=5 \times 5=25$A×A=5×5=25 Divide 25 order pairs of $A \times A$A×A in 3 parts as follows : $$\begin{aligned} & \text { Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9) \\ & \text { Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6), \\ & (4,8),(4,9),(6,8),(6,9),(8,9) \\ & \text { Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), } \\ & (9,4),(8,6),(9,6),(9,8) \end{aligned}$$​ 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 $-A$−A, both components of each order pair are same. In part $-B$−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 $-C$−C, only reverse of the order pairs of part $-B$−B are present i.e., if $(a, b)$(a,b) is present in part $-B$−B, then (b, a) will be present in part $-C$−C For example $(3,4)$(3,4) is present in part $-B$−B and $(4,3)$(4,3) present in part $-C$−C. Number of order pair in $A, B$A,B and $C$C are 5, 10 and 10 respectively. In any symmetric relation on set $A$A, if any order pair of part $-B$−B is present then its reverse order pair of part $-C$−C will must be also present. Hence number of symmetric relation on set $A$A is equal to the number of all relations on a set $D$D, which contains all the order pairs of part $-A$−A and part $-B$−B. Now $n(D)=n(A)+n(B)=5+10=15$n(D)=n(A)+n(B)=5+10=15 Hence number of all relations on set $D=(2)^{15}$D=(2)15 $\Rightarrow$⇒ Number of symmetric relations on set $D=(2)^{15}$D=(2)15