We are given that n(A × B) = 9. Since n(A × B) = n(A) × n(B) and 9 is a perfect square (3 × 3), it implies n(A) = 3 and n(B) = 3 (if A=B). In a Cartesian product A × B, the first element of each ordered pair belongs to set A and the second element belongs to set B. From the pair (-1, 0), we know -1 ∈ A and 0 ∈ B. From the pair (0, 1), we know 0 ∈ A and 1 ∈ B. Since the total number of elements in A must be 3 to satisfy n(A) × n(B) = 9, and we have identified -1 and 0 as elements of A from the given pairs, the elements of A must be {-1, 0, 1} assuming A = B as per standard NCERT textbook problems of this type. Even if A was not equal to B, the first elements of the pairs provided are -1 and 0; however, to reach a product of 9, set A must have 3 elements. The only option containing 3 elements that fits the observations is {-1, 0, 1}. Options A, C, and D only contain 2 elements, which would make n(A × B) either 4 or 6, not 9.