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: \(\boxed{2}\)
Five friends A, B, C, D, and E are sitting in a row facing north, but not necessarily in the same order:
B is to the immediate left of C
E is not at any of the ends
D is to the right of E but not next to C
A is at one of the ends
Who is sitting in the middle?
If \((2m+n) + (2n+m)=27\), find the maximum value of \((2m-3)\), assuming m and n are positive integers.