Read the following information and answer the question that follows. P, Q, R,…

Concept: In a circular seating puzzle where everyone faces the centre, “second to the left” and similar directional clues move in the clockwise direction as seen from above (facing the centre puts the clockwise arc on a person’s left). Fix one person as a reference seat, convert each clue into a seat offset from that reference, and resolve the remaining seats using the adjacency clues that are left.

1. Fix S at seat 0. From clue (ii), V is second to S’s left, i.e. 2 seats clockwise from S, so V sits at seat 2.

2. V’s neighbours are T and U (clue ii), so seats 1 and 3 hold T and U in some order.

3. From clue (iii), S is not a neighbour of T. S’s own neighbours are seat 6 and seat 1, so seat 1 cannot be T — seat 1 = U and seat 3 = T.

4. The remaining seats 4, 5 and 6 hold P, Q and R. From clue (iv), P is a neighbour of both Q and R, so P must take the middle seat of that block: seat 5 = P, with Q and R on seats 4 and 6.

5. From clue (iii), S is also not a neighbour of R. S’s other neighbour is seat 6, so seat 6 cannot be R — seat 6 = Q and seat 4 = R.

6. The resolved clockwise order is: S(0) – U(1) – V(2) – T(3) – R(4) – P(5) – Q(6) – back to S.

7. Now interchange S and T: T moves into seat 0 and S moves into seat 3. S’s new neighbours are seat 2 (V) and seat 4 (R).

Cross-check: once S and T merely swap seats, S simply inherits the seat T used to occupy, so S’s new neighbours are exactly T’s original neighbours. T (seat 3) originally sat between V (seat 2) and R (seat 4) — confirming the same result independently.

So after T and S interchange their places, S becomes a neighbour of V and R.