A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done, if the committee should have 4 professors and 1 research associate or all 3 trainees and 2 professors?

### Concept & Addition Principle Use combinations $^nC_r$ to select group members. The word "and" indicates multiplication between selections, while "or" indicates addition between separate, mutually exclusive cases. ### Step-by-Step Solution 1. **Case 1: 4 professors AND 1 research associate.** Select 4 professors out of 4: $^4C_4 = 1$ way. Select 1 research associate out of 6: $^6C_1 = 6$ ways. Ways for Case 1 = $1 \times 6 = 6$. 2. **Case 2: 3 trainees AND 2 professors.** Select 3 trainees out of 3: $^3C_3 = 1$ way. Select 2 professors out of 4: $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$ ways. Ways for Case 2 = $1 \times 6 = 6$. 3. **Total Ways (Case 1 OR Case 2):** Total = $6 + 6 = 12$ ways. ### Exam Strategy & Shortcut Recognize that selecting all members of a specific subgroup (like 4 professors out of 4) is always just 1 way ($^nC_n = 1$). This immediately simplifies the multiplication to $1 \times 6$ for both cases. ### Common Pitfall Multiplying the results of Case 1 and Case 2 instead of adding them. Remember: OR means ADD. ### Final Answer Therefore, the correct answer is **12**.