Practicing Success
For any three sets A, B and C the set $(A∪B∪C)∩(A∩B'∩C')'∩C'$ is equal to |
$B∩C'$ $B'∩C'$ $B∩C$ $A∩B∩C$ |
$B∩C'$ |
The correct answer is Option (1) → $B∩C'$ We have, $(A∩B'∩C')' = A'∪B∪C$ $∴(A∪B∪C)∩(A∩B'∩C')'$ $=(A∪B∪C)∩(A'∪B∪C)$ $=(A∩A')∪(B∪C)$ [By distributivity ∪ of over ∩] $=\phi∪(B∪C) = B∪C$ Hence, $(A∪B∪C)∩(A∩B'C')' ∪C'$ $=(B∪C)∩C' = (B∩C')∪(C∩C') = B∩C'$ |