KCET · Maths · Sets and Relations
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 \(5^2\)
- D \(2^5\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
Let \(n(A) = p\) and \(n(B) = q\).
The number of relations from set A to set B is given by \(2^{p \times q}\).
Given \(n(A) = 2\), we have \(p = 2\).
The number of relations is \(1024\), so \(2^{2q} = 1024\).
Since \(1024 = 2^{10}\), we get \(2^{2q} = 2^{10}\).
Equating the exponents, \(2q = 10 \Rightarrow q = 5\).
Therefore, \(n(B) = 5\).
Answer: \(5\)
The number of relations from set A to set B is given by \(2^{p \times q}\).
Given \(n(A) = 2\), we have \(p = 2\).
The number of relations is \(1024\), so \(2^{2q} = 1024\).
Since \(1024 = 2^{10}\), we get \(2^{2q} = 2^{10}\).
Equating the exponents, \(2q = 10 \Rightarrow q = 5\).
Therefore, \(n(B) = 5\).
Answer: \(5\)
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