$\textbf{For symmetric relation:}$ We can also think of it as a matrix $n\times n$, with the elements of the matrix being $(a_i,a_j)$ with $ a_i,a_j \in A$. The elements of the main diagonal can be perfectly chosen for the relation because they are symmetric. For the rest of the elements, picking a pair from the upper triangle say $(a_2,a_1)$ implies that you are also picking $(a_1,a_2)$. So from the total $n^2$ pairs you end up with only $$\sum_{i=0}^{n} i = \frac{n(n+1)}{2}$$ from which to choose. We can do this in $2^{\frac{n(n+1)}{2}}$ ways.
So, $n(S) = 2^{\frac{n(n+1)}{2}} = 2^{10} = 1024$
$\textbf{For asymmetric relation:}$ Let us take an example.
$$A = \{1,2,3\}$$
$$A\times A = \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$$
We know that, $(1,1),(2,2),(3,3)$ won't come in the asymmetric relation.(3 elements)
Remaining elements of $A\times B$ can be grouped into 3 groups of 2 elements each as:
$[(1,2),(2,1)]$, $[(1,3),(3,1)]$, $[(2,3),(3,2)]$ (3 groups)
From every group, we can choose elements in 3 different ways.
1. choose 1st element.
2. choose 2nd element.
3. choose no elements.
That is, we can choose elements in $3^3$ different ways in this case.
Generalizing,
If there are $n$ elements in the set, the number of elements in the group is $\frac{\text{Total number of relations - reflexive elements }}{2}=\frac{n^2 - n}{2}$
Each of these elements can be chosen in 3 ways. Therefore the number of asymmetric relations is $3^{\frac{n^2 - n}{2}}.$
So, $n(A) = 3^{\frac{n^2 - n}{2}} = 3^{6} = 729$
Some examples:
$R_{1}=\{(1,1),(1,2),(2,1)\}$, it is symmetric but not asymmetric.
$R_{2}=\{(1,2)\}$, it is asymmetric but not symmetric.
$R_{3}=\{\:\:\}$ , it is symmetric and asymmetric.
$R_{4}=\{(1,1),(1,2)\}$, it is not symmetric and not asymmetric.
$n(S\cap A) = 1$ because empty relation is the only one which is both symmetric as well as asymmetric.
Now, $n(S\cup A)=n(S) + n(A) - n(S \cap A)$
$\implies n(S \cup A) = 1024 + 729 - 1 = 1752$.
So, the correct answer is $1752$.
