We know that
(A×B)∩(C×D)=(A∩C)×(B∩D)
∴(A × B)∩(B × A) = (A∩B) × (B∩A)
⇒(A × B)∩(B × A) = (A∩B) × (A∩B)
It is given that A∩B has n elements.
∴(A∩B) × (A∩B) has n^2 elements.
But, (A× B)∩(B × A) = (A∩B) × (A∩B)
∴(A× B)∩(B × A) has n^2 elements in common. |