Which of the following combinations of test results is NOT possible?

The correct answer/option is: "Vials A and E positive, vials C and D negative".

To understand why this combination of test results is not possible, we can analyze the distribution of blood samples for the 16 patients (numbered 1 through 16) into the 8 vials (labeled A through H).

The distribution table of blood samples is as follows:
• Vial A: Patients 9, 10, 11, 12, 13, 14, 15, 16
• Vial B: Patients 1, 2, 3, 4, 5, 6, 7, 8
• Vial C: Patients 5, 6, 7, 8, 13, 14, 15, 16
• Vial D: Patients 1, 2, 3, 4, 9, 10, 11, 12
• Vial E: Patients 3, 4, 7, 8, 11, 12, 15, 16
• Vial F: Patients 1, 2, 5, 6, 9, 10, 13, 14
• Vial G: Patients 2, 4, 6, 8, 10, 12, 14, 16
• Vial H: Patients 1, 3, 5, 7, 9, 11, 13, 15

Key Principle of Group Testing:
1. Exactly one patient has the disease.
2. If a patient is infected, all vials containing their blood sample must test positive.
3. If a vial tests negative, it means the infected patient is not among the patients whose blood was added to that vial.

Let us examine the relationship between Vials C and D:
• The set of patients in Vial C is {5, 6, 7, 8, 13, 14, 15, 16}.
• The set of patients in Vial D is {1, 2, 3, 4, 9, 10, 11, 12}.

Notice that these two sets are mutually exclusive and collectively exhaustive. That is, every patient from 1 to 16 belongs to either Vial C or Vial D, but not both:

$$\text{Vial C}\cup \text{Vial D}=\{1,2,3,...,16\}$$

Since exactly one patient among the 16 is diseased, the infected patient must be in either Vial C or Vial D. Therefore, it is mathematically mandatory that at least one of Vials C or D tests positive.

If both Vials C and D test negative, it would imply that none of the 16 patients has the disease, which directly contradicts the premise that exactly one patient is infected.

Thus, the combination where both vials C and D are negative is logically impossible.