A relation on a set $A$ is said to be symmetric iff $\left( {a,b} \right) \in A \Rightarrow \left( {b,a} \right) \in A,\forall a,b \in A$

Here $A = \left\{ {3,4,6,8,9} \right\}$

Number of order paris of $A \times A = 5 \times 5 = 25$

Divide $25$ order pairs of $A \times A$ in 3 parts as follows:

Part- $A:\left( {3,3} \right),\left( {4,4} \right),\left( {6,6} \right),\left( {8,8} \right),\left( {9,9} \right)$

Part- $B:\left( {3,4} \right),\left( {3,6} \right),\left( {3,8} \right),\left( {3,9} \right),\left( {4,6} \right),\left( {4,8} \right),\left( {4,9} \right),\left( {6,8} \right),\left( {6,9} \right),\left( {8,9} \right)$

Part- $C:\left( {4,3} \right),\left( {6,3} \right),\left( {8,3} \right),\left( {9,3} \right),\left( {6,4} \right),\left( {8,4} \right),\left( {9,4} \right),\left( {8,6} \right),\left( {9,6} \right),\left( {9,8} \right)$

In part – $A$ , both components of each order pair are same.

In part – $B$ , both components are different but not two such order paris are present in which fiest 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 peresent i.e., if $(a,b)$ is persent in part – $B$, then $(b,a)$ will be present in part – $C$ For example $(4,3)$ is peresnt in part – $C$ .

Number of order pair in $A$,$B$ and $C$ are $5,10$ and $10$ respectively.

In any symmetic 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 aset $D$, which contains all the order pairs of part – $A$ and part – $B$.

Now $n\left( D \right) = n\left( A \right) + n\left( B \right) = 5 + 10 = 15$

Hence number of all relations on set $D = {\left( 2 \right)^{15}}$

$ \Rightarrow $ Number of symmetric relations on set $D = {\left( 2 \right)^{15}}$