1. **Problem statement:** We have 12 students in a class, but one student is only there for refreshment and does not participate in experiments. So, effectively, 11 students are available for grouping. 2. The students do experiments in groups of 3. 3. Mr Tau and Miss Julia do not want to be in the same group. 4. We need to find the number of possible groups of 3 students that can be formed such that Mr Tau and Miss Julia are not in the same group. 5. **Step 1: Calculate total groups of 3 from 11 students.** The number of ways to choose 3 students from 11 is given by the combination formula: $$\binom{11}{3} = \frac{11!}{3! \times (11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$$ 6. **Step 2: Calculate groups that include both Mr Tau and Miss Julia.** If both Mr Tau and Miss Julia are in the group, we need to choose 1 more student from the remaining 9 students (since 11 total minus these 2): $$\binom{9}{1} = 9$$ 7. **Step 3: Calculate groups that do NOT include both Mr Tau and Miss Julia together.** This is total groups minus groups that include both: $$165 - 9 = 156$$ 8. **Answer:** There are **156** possible groups of 3 students that do not include both Mr Tau and Miss Julia together.