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Chapter 1: Problem 23

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

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

Step by step solution

01

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: \[B \times A = \{(b, a) \mid b \in B \text{ and } a \in A\}\]
02

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: \[(A \times B) \cap (B \times A) = \{(x, y) \mid (x, y) \in (A \times B) \text{ and } (x, y) \in (B \times A)\}\]
03

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).
04

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.

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Most popular questions from this chapter

If \(\mathrm{U}=\\{1,2,3\\}\) and \(\mathrm{A}=\\{1,2\\}\) then \([\mathrm{P}(\mathrm{A})]^{\prime}=\ldots \ldots \ldots\) (a) \(\\{\\{3\\},\\{2,3\\},\\{1,3\\},\\{1,2\\}, \Phi\\}\) (b) \(\\{\\{3\\},\\{2,3\\},\\{1,3\\},\\{1,2,3\\}\\}\) (c) \(\\{\\{3\\},\\{2,3\\},\\{1,3\\},\\{1,2,3\\}, \Phi\\}\) (d) \(\\{\\{3\\},\\{2,3\\},\\{1,3\\},\\{1,2\\}\\}\)

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