1. **Problem statement:** There are 10 members sitting around a round table, including 1 chairman seat. Among them, 4 members form a written report subcommittee. We want to find the number of ways to seat all 10 members around the table under two conditions: a. The 4 subcommittee members must sit together. b. No two subcommittee members sit next to each other. --- 2. **Key concepts and formulas:** - For seating $n$ distinct people around a round table, the number of distinct arrangements is $(n-1)!$ because rotations are considered the same. - When a group must sit together, treat them as a single block, then multiply by the internal arrangements of that block. - When no two members of a group can sit together, use combinatorial methods to place them with gaps. --- ### Part (a): Subcommittee members sit together 3. Treat the 4 subcommittee members as a single block. Along with the other 6 members, we have $6 + 1 = 7$ entities to arrange around the table. Number of ways to arrange these 7 entities around a round table: $$ (7 - 1)! = 6! = 720 $$ 4. The 4 subcommittee members inside their block can be arranged among themselves in: $$ 4! = 24 $$ 5. Total number of ways: $$ 6! \times 4! = 720 \times 24 = 17280 $$ --- ### Part (b): No two subcommittee members sit next to each other 6. First, seat the 6 non-subcommittee members around the table. Number of ways: $$ (6 - 1)! = 5! = 120 $$ 7. These 6 members create 6 gaps around the table where the 4 subcommittee members can be placed so that no two are adjacent. Number of ways to choose 4 gaps out of 6: $$ \binom{6}{4} = 15 $$ 8. Arrange the 4 subcommittee members in the chosen gaps: $$ 4! = 24 $$ 9. Total number of ways: $$ 5! \times \binom{6}{4} \times 4! = 120 \times 15 \times 24 = 43200 $$