Six persons A, B, C, D, E and F are sitting around a circular table facing the…

Concept: For circular seating puzzles where everyone faces the centre, first fix a numbering direction (say clockwise) and note that facing inward reverses left/right relative to that fixed direction — moving clockwise is each person's own left, and moving anticlockwise is each person's own right. Convert every clue into an offset using this rule and place people step by step; on a table of 6 seats, the seat diagonally opposite any given seat is exactly 3 seats away (half of 6) in either direction.

Application:

1. Fix F at seat 1 (numbering the seats 1 to 6 clockwise). Since everyone faces the centre, "right" means anticlockwise here — Statement III places D two seats to F's right, i.e. two seats anticlockwise of seat 1, which is seat 5. So D takes seat 5.

2. Statement I places C between A and F, so C must be adjacent to F on one side and to A on the other — giving a running sequence of A-C-F or F-C-A around the circle.

3. Testing C at seat 6 (F's anticlockwise neighbour) would force A onto seat 5 to complete the sequence, but seat 5 is already taken by D — a contradiction, so this branch is rejected.

4. Testing C at seat 2 (F's clockwise neighbour) instead works: A takes seat 3 to complete F(1)-C(2)-A(3), with no clash against D at seat 5.

5. Only seats 4 and 6 remain unfilled. Since everyone faces the centre, "left" means clockwise here — Statement II places B two seats to E's left, i.e. two seats clockwise of E; putting E at seat 4 gives B at seat 4 + 2 = seat 6 — exactly the two seats left over.

Cross-check: re-reading all three statements against this arrangement confirms each holds — C(2) sits between A(3) and F(1); D(5) is two seats anticlockwise of (to the right of) F(1); B(6) is two seats clockwise of (to the left of) E(4). Since the table has 6 seats, the seat diagonally opposite A's seat (3) is seat 3 + 3 = 6 — occupied by B.

So B is sitting diagonally opposite A.