If A=\ x ∈ z^ + : x < 10. and x is a multiple of 3 or 4\ , where z^...

A relation on a set A is said to be symmetric iff $(a, b) \in A \Rightarrow(b, a) \in A, \forall a, b \in A$ Here A=\{3,4,6,8,9\}Number of $order pairs of A \times A=5 \times 5=25Divide 25 order pairs of$A \times A in$3 parts as follows :$$\begin{aligned} & \text{ Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9) \\ & \text{ Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6), \\ & (4,8),(4,9),(6,8),(6,9),(8,9) \\ & \text{ Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), } \\ & (9,4),(8,6),(9,6),(9,8) \end{aligned}$$$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} \Rightarrow$Number of symmetric relations on set$D=(2)^{15}$