Let's break down the problem step by step.

Given:

• $n(P) = 3$ (number of elements in set $P$)
• $n(Q) = 5$ (number of elements in set $Q$)
• $n(PQ^c \cup U) = 2$ (where $U$ is the universal set)

We need to find $n((P \times Q) \cap (Q \times P))$.

Step 1: Understand $P \times Q$ and $Q \times P$

• The Cartesian product $P \times Q$ consists of all ordered pairs $(p, q)$ where $p$ is an element of $P$ and $q$ is an element of $Q$. The total number of elements in $P \times Q$ is $n(P) \times n(Q) = 3 \times 5 = 15$.

• Similarly, $Q \times P$ consists of all ordered pairs $(q, p)$ where $q$ is an element of $Q$ and $p$ is an element of $P$. The total number of elements in $Q \times P$ is also $n(Q) \times n(P) = 5 \times 3 = 15$.

Step 2: Find $(P \times Q) \cap (Q \times P)$

• The intersection $(P \times Q) \cap (Q \times P)$ will contain pairs that are in both $P \times Q$ and $Q \times P$. This means the pair $(p, q)$ from $P \times Q$ must be equal to $(q, p)$ from $Q \times P$.

• For $(p, q) = (q, p)$ to hold, $p$ must be equal to $q$. Therefore, the elements in $P$ and $Q$ that contribute to this intersection must be common elements between $P$ and $Q$.

• The number of such common elements will determine the size of $(P \times Q) \cap (Q \times P)$.

Step 3: Determine $n((P \times Q) \cap (Q \times P))$

• Given that $n(P) = 3$ and $n(Q) = 5$, and there is no information provided about the specific elements in $P$ and $Q$, the worst-case assumption is that no elements are common between $P$ and $Q$.

• If $P$ and $Q$ have no common elements, $n((P \times Q) \cap (Q \times P)) = 0$.

However, if we consider the possibility of at least one element being common, then the minimum possible value would be when one element is common between $P$ and $Q$, giving us $1 \times 1 = 1$ pair.

However, the problem setup seems more inclined towards a trick question. Let's carefully analyze the options provided, assuming that one common element exists and maximum interaction:

Conclusion:

Given this scenario and the setup provided, the answer is likely to be the case where the sets $P$ and $Q$ have 3 common elements, since this would give the largest non-trivial intersection. This leads to:

• $n((P \times Q) \cap (Q \times P)) = 3^2 = 9$.

So, the correct answer is $\boxed{9}$.

Do you want more details or have any questions? Here are some relative questions:

1. How are Cartesian products defined in set theory?
2. What happens when sets $P$ and $Q$ have no common elements?
3. Can $(P \times Q) \cap (Q \times P)$ ever be larger than $n(P) \times n(Q)$?
4. How does the universal set $U$ relate to the problem?
5. What implications arise if all elements in $P$ are found in $Q$?

Tip: Understanding how intersections work in Cartesian products is crucial for set theory problems.