Question

If $n(A) = 2$ and the number of relations from set A to set B is 1024, then $n(B)$ is

• A. 2
• B. 5
• C. $2^5$
• D. $5^2$

Hint:

It is very beneficial to memorize the powers of 2 up to $2^{10} = 1024$ for competitive exams. It speeds up problems involving subsets, combinations, and binary calculations significantly.

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between sets and the concept of relations between two sets.

Given, \(n(A) = 2\). It means set A has 2 elements.

The number of relations from set A to set B is given as 1024. A relation from set A to set B can be defined as a subset of the Cartesian product \(A \times B\).

The number of different relations from set A to set B is given by the formula \(2^{n(A) \times n(B)}\).

We need to determine \(n(B)\), the number of elements in set B.

According to the information provided, the number of relations is 1024:

\(2^{n(A) \times n(B)} = 1024\)

We know that \(1024 = 2^{10}\). Thus, comparing the exponents: \(n(A) \times n(B) = 10\).

Since \(n(A) = 2\), we substitute:

\(2 \times n(B) = 10\)

Solving for \(n(B)\):

\(n(B) = \frac{10}{2} = 5\)

Therefore, the number of elements in set B is \(5\). Hence, the correct answer is option: 5.