If P and Q be the two sets such that n(P) = 3 and n (Q) = 5, if n(PQ0U) = 2, then n ((P × Q) ∩ (Q × P)) equals. (where U is the universal set)

(1) 15

(2) 9

(3) 25

(4) 4

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.

Explore how to calculate the intersection of Cartesian products in set theory. Understand the concepts of set intersections, Cartesian products, and their applications. Suitable for advanced high school and college students.

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