A is CORRECT. We saw this scenario in answer E of question 14:
It’s an easy diagram to make because M and K are both random.
B and C don’t work because they don’t have more people in group 3 than in group 2.
For B, if T is in group 3, then SU must be in group 2. So each group has two.
For C, if P is in group 3 then T must be in group 1 and SU are automatically in group 2. So groups 2 and 3 each have two people.
(These deductions are based on rules 1 and 2)
D and E violate either the first or second rule. Two our of three of P, T and U must be in groups 1 and 2. Both of these answers place too many of those variables in group 3.
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