If $$A = \{x \in z^+ : x < 10$$ 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

The set $$A$$ is defined as all positive integers $$\lt 10$$ that are multiples of $$3$$ or $$4$$.

Multiples of $$3$$ below $$10$$: $$3, 6, 9$$.
Multiples of $$4$$ below $$10$$: $$4, 8$$.

Taking the union and removing repetition we get
$$A = \{3,4,6,8,9\}$$, so $$|A| = 5$$.

A relation on $$A$$ is a subset of $$A \times A$$. For the relation to be symmetric, whenever $$(a,b)$$ is present, $$(b,a)$$ must also be present.

Counting choices for a symmetric relation on an $$n$$-element set:

• Diagonal (reflexive) ordered pairs $$(a,a)$$: there are $$n$$ of them. Each can be either included or not, giving $$2^n$$ choices.

• Off-diagonal ordered pairs: group them into unordered pairs $$\{a,b\}$$ with $$a \neq b$$. Each unordered pair corresponds to the two ordered pairs $$(a,b)$$ and $$(b,a)$$ which must be chosen together (both in or both out). The number of such unordered pairs is $$\binom{n}{2} = \frac{n(n-1)}{2}$$, giving $$2^{\frac{n(n-1)}{2}}$$ choices.

Therefore, the total number of symmetric relations is
$$2^n \times 2^{\frac{n(n-1)}{2}} = 2^{\,n + \frac{n(n-1)}{2}} = 2^{\frac{n(n+1)}{2}}$$.

For $$n = 5$$, this exponent equals
$$\frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 15$$.

Hence the number of symmetric relations on $$A$$ is $$2^{15}$$.

Option B which is: $$2^{15}$$