(i) B ∩ C = { }
∴ A x (B ∩ C) = φ ………….. (1)
A x B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2,4)}
A x C = {(1, 5), (1,6), (2, 5) (2,6)}
∴ (A x B) ∩ (A x C) = φ ………………. (2)
From (1) and (2), we get
A x (B∩C) = (A x B) ∩(A x C)
(ii) A x C = {(1, 5), (1,6), (2, 5), (2, 6)}
B x D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}.
Clearly every elements of A x C is an element of B x D. A x C ⊂B x D.