A, B, C, D, E, F and G are sitting around a circular table facing towards its…

For a circular arrangement where everyone faces the centre, the everyday left/right sense is reversed compared with facing outward: a seated person's right hand points in the anticlockwise direction and their left hand points in the clockwise direction, so a clue such as "X sits n-th to the right/left of Y" fixes X exactly n seats anticlockwise/clockwise of Y, and "immediate right/left neighbour" means one seat away in that same direction. Solve such puzzles by anchoring one person's seat as a reference, converting every directional clue into a seat offset from that anchor, and testing the seats still open against any leftover clues -- including exclusions such as "is not a neighbour of" -- until exactly one arrangement of everybody is consistent with every clue.

1. Fix C's seat as the reference point, seat 0, around the seven-seat circle (seats numbered clockwise).

2. "E sits in the immediate right-hand neighbourhood of C" places E one seat anticlockwise of C, i.e. seat 0 minus 1 = seat 6.

3. "C sits third to the left of D" places C three seats clockwise of D, so D sits three seats anticlockwise of C, i.e. seat 0 minus 3 = seat 4 (wrapping round the seven-seat circle).

4. "E sits third to the right of F" places E three seats anticlockwise of F, so F sits three seats clockwise of E, i.e. seat 6 plus 3 = seat 9, which wraps to seat 2.

5. C, E, D and F now occupy seats 0, 6, 4 and 2, leaving seats 1, 3 and 5 for A, B and G.

6. "Counting from the left of B, only one person sits between G and B" needs G two seats clockwise of B (clockwise is B's own left). Of the leftover seats, only B at seat 1 with G at seat 3 satisfies this, with F at seat 2 as the single person between them; pairing B at seat 3 would force G onto seat 5 and A onto seat 1, which the next clue rules out.

7. That leaves A at seat 5. Checking "A is not an immediate neighbour of C" (seat 0): C's neighbours are seats 6 and 1, and seat 5 is neither, so this arrangement satisfies every clue.

Cross-check: going the other way around -- anticlockwise, i.e. towards A's right, from A's seat 5 towards C's seat 0 -- passes through seats 4, 3, 2 and 1 (D, G, F and B), four people; since A and C themselves take up two of the seven seats, the two arcs between them must together hold the other five people, and 4 (the long way, via A's right) plus 1 (the short way, via A's left) accounts for all five. Counting from A's left (clockwise) side, as the question asks, the only seat passed before reaching C is seat 6, held by E -- so exactly one person sits between A and C, and the answer is "one".