1. **Problem statement:** We have a committee of 6 people sitting in a row. Two specific members refuse to sit next to each other. We need to find the number of possible seating orders under this restriction. 2. **Total number of ways without restriction:** The total number of ways to arrange 6 people in a row is given by the factorial of 6: $$6! = 720$$ 3. **Number of ways where the two members sit together:** Treat the two members who refuse to sit together as a single unit. Then we have this unit plus the other 4 members, making 5 units total. Number of ways to arrange these 5 units: $$5! = 120$$ The two members inside the unit can be arranged in: $$2! = 2$$ So, total arrangements with the two members together: $$5! \times 2! = 120 \times 2 = 240$$ 4. **Number of ways where the two members do NOT sit together:** Subtract the number of arrangements where they sit together from the total arrangements: $$6! - (5! \times 2!) = 720 - 240 = 480$$ 5. **Final answer:** There are **480** possible seating orders where the two members do not sit next to each other.