Since this is a circular arrangement, we fix A at the top position to eliminate rotational symmetry. This simplifies counting without affecting the final answer.
Step 1: Place B relative to A.
B sits third to the left of A. Since everyone faces the center, left means counter-clockwise. So B is placed three seats counter-clockwise from A.
Step 2: Use the restriction involving D and F.
F sits immediately to the right of D. Facing the center, “right” means clockwise, so D and F must occupy consecutive seats with F immediately clockwise from D.
Step 3: Apply the condition between C and D.
Only one person sits between C and D. Therefore, C must be exactly two seats away from D, and this can occur in two ways:
- C is two seats clockwise from D, or
- C is two seats counter-clockwise from D.
These two cases must be tested separately.
Step 4: Apply the restriction on E.
E is not a neighbor of A or C. After placing A, B, D, F, and C in each possible case, only two seats remain. E must go into the seat that is not adjacent to A or C. This eliminates invalid possibilities.
Final Conclusion:
After testing all allowed placements systematically, exactly two circular arrangements satisfy all the given seating conditions.
Final Answer: \(\boxed{2}\)
Four teams – Red (R), Blue (B), Green (G), and Yellow (Y) – are competing in the final four rounds of the Inter-School Science Olympiad, labeled Round A, Round B, Round C, and Round D. Each round consists of one match between two teams, and every team plays exactly two matches. No team plays the same opponent more than once.
The final schedule must adhere to the following rules:
(193 words)
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: