Sets of three integers with LCM 1001, HCF 1

The prime factorization of 1001 = 7 × 11 × 13. For a set of three numbers \{a, b, c\}, the LCM dictates the maximum exponent of each prime factor must be 1, and the HCF dictates the minimum exponent must be 0. For each prime, the valid exponent triplets (e_a, e_b, e_c) that contain at least one 1 and at least one 0 are: (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). This gives 6 ordered arrangements per prime. Total ordered triplets = 6 × 6 × 6 = 216. To find unordered sets of distinct numbers, we remove cases where two numbers are identical (24 cases) and divide by 3! (6 arrangements). Unordered sets = (216 - 24) / 6 = 32. Since 32 is strictly greater than 8, the answer is "More than 8".

Answer: (d).