Question

If \(\mathrm{n}(\mathrm{A})=3, \mathrm{n}(\mathrm{B})=5\) and \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=2\) then \(\mathrm{n}[(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})]=\ldots \ldots\) (a) 5 (b) 3 (c) 4 (d) 6

Short answer

The number of elements in the intersection of the Cartesian products A x B and B x A is 4.

Step-by-step solution

Understand the Cartesian product

The Cartesian product of two sets A and B, written as A x B, is the set of all ordered pairs (a, b) where a is from A and b is from B. We can write it as: \[A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}\] Similarly, the Cartesian product of B x A is the set of all ordered pairs (b, a) where b is from B and a is from A. We can write it as: \[B \times A = \{(b, a) \mid b \in B \text{ and } a \in A\}\]

Find the intersection of Cartesian products

To find the intersection of the Cartesian products A x B and B x A, we need to find the set of all ordered pairs that are present in both sets. This can be represented as: \[(A \times B) \cap (B \times A) = \{(x, y) \mid (x, y) \in (A \times B) \text{ and } (x, y) \in (B \times A)\}\]

Analyze the intersection

Notice that for an ordered pair (x, y) to be in both A x B and B x A, x must be from both A and B, and y must also be from both A and B. This implies that x and y are from the intersection of A and B. In other words, (x, y) must be an element of (A ∩ B) x (A ∩ B).

Count the elements in the intersection

We are given that n(A ∩ B) = 2, which means there are 2 elements in the intersection of A and B. To get the total number of ordered pairs in the Cartesian product (A ∩ B) x (A ∩ B), we multiply the number of elements in A ∩ B by itself (since there are 2 elements in both A ∩ B and B ∩ A). \[\text{n}[(A \times B) \cap (B \times A)] = \text{n}[(A \cap B) \times (A \cap B)] = 2 \times 2 = 4\] So, the correct answer is (c) 4.

Key concepts

Cartesian Product

The Cartesian product is a fundamental concept in set theory that involves creating pairs from elements of two sets. When we take the Cartesian product of sets \( A \) and \( B \), denoted as \( A \times B \), we form a new set. This set contains all possible ordered pairs \( (a, b) \), where \( a \) is an element from set \( A \), and \( b \) is an element from set \( B \).

For example, if set \( A = \{1, 2\} \) and set \( B = \{' x', 'y'\} \), then the Cartesian product \( A \times B \) will be \( \{ (1, 'x'), (1, 'y'), (2, 'x'), (2, 'y') \} \).

Keep in mind that the Cartesian product of \( B \times A \) is not identical to \( A \times B \), as the order of the elements matters. For \( B \times A \), the set will contain \( ('x', 1), ('x', 2), ('y', 1), ('y', 2) \). Notice how each pair has switched places compared to \( A \times B \).

Intersection of Sets

When we talk about the intersection of sets, we're looking for elements that are common in two or more sets. The intersection of sets \( A \) and \( B \), denoted as \( A \cap B \), includes only those elements that are in both set \( A \) and set \( B \).

Suppose \( A = \{1, 2, 3\} \) and \( B = \{2, 3, 4\} \). The intersection \( A \cap B \) will be \( \{2, 3\} \) because 2 and 3 are the elements present in both sets.

In the problem we're solving, \( n(A \cap B) = 2 \) tells us that there are two common elements in \( A \) and \( B \). When extending this concept to Cartesian products, the intersection \( (A \times B) \cap (B \times A) \) involves finding ordered pairs present in both Cartesian products. Here, it simplifies to finding the Cartesian product of the intersection set \( (A \cap B) \times (A \cap B) \).

Ordered Pairs

Ordered pairs are a basic concept in mathematics used to define elements in a Cartesian product. An ordered pair is a set of two elements \( (x, y) \), where the order of the elements matters. This is crucial because \( (x, y) \) is not the same as \( (y, x) \) unless \( x = y \).

Using ordered pairs, we can represent points in a plane, coordinates for mapping, and other mathematical constructs. In our problem, the ordered pairs arising from the Cartesian products \( A \times B \) and \( B \times A \) need to be carefully analyzed, especially when finding common pairs in their intersection.

For ordered pairs to exist in both \( A \times B \) and \( B \times A \), the components of the pairs must be elements from \( A \cap B \). Therefore, we derive that each pair \( (x, y) \) within this intersection is formed from elements that are present in both \( A \) and \( B \). This understanding helps solve problems involving intersections of Cartesian products more efficiently.