Question:
Match List-I with List-II. Suppose \(A, B, C\) are three non-empty sets.
List-I:
A. De-Morgan's Law,
B. Associative Law,
C. Distributive Law,
D. Difference of sets Law.
Match List-I with List-II. Suppose \(A, B, C\) are three non-empty sets.
List-I:
A. De-Morgan's Law,
B. Associative Law,
C. Distributive Law,
D. Difference of sets Law.
Show Hint
Set difference \(A-B\) can always be written as \(A\cap B^c\).
Updated On: Jun 8, 2026
- A-I, B-II, C-III, D-IV
- A-I, B-IV, C-III, D-II
- A-II, B-IV, C-III, D-I
- A-IV, B-III, C-II, D-I
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
This question is based on standard laws of sets.
Step 1: De-Morgan's Law. \[ (A\cup B)^c=A^c\cap B^c \] So, \[ A\rightarrow II \]
Step 2: Associative Law. \[ A\cup(B\cup C)=(A\cup B)\cup C \] So, \[ B\rightarrow IV \]
Step 3: Distributive Law. \[ A\cup(B\cap C)=(A\cup B)\cap(A\cup C) \] So, \[ C\rightarrow III \]
Step 4: Difference of sets Law. \[ A\setminus B=A\cap B^c \] So, \[ D\rightarrow I \] Therefore, \[ A-II,\ B-IV,\ C-III,\ D-I \] \[ \therefore \text{Correct Answer is (C)} \]
This question is based on standard laws of sets.
Step 1: De-Morgan's Law. \[ (A\cup B)^c=A^c\cap B^c \] So, \[ A\rightarrow II \]
Step 2: Associative Law. \[ A\cup(B\cup C)=(A\cup B)\cup C \] So, \[ B\rightarrow IV \]
Step 3: Distributive Law. \[ A\cup(B\cap C)=(A\cup B)\cap(A\cup C) \] So, \[ C\rightarrow III \]
Step 4: Difference of sets Law. \[ A\setminus B=A\cap B^c \] So, \[ D\rightarrow I \] Therefore, \[ A-II,\ B-IV,\ C-III,\ D-I \] \[ \therefore \text{Correct Answer is (C)} \]
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