Six families A, B, C, D, E and F are living in houses in a row. B has F and D as neighbours, E has A and C as neighbours, A does not live next to either F or D. C does not live next to D. Who are F’s next door neighbours?

Work it as a block problem. B has F and D on either side, so F-B-D forms one block. E has A and C on either side, so A-E-C forms a second block. Six families in six houses means the two blocks fill the row end to end.

The constraints decide the orientation. A must not sit next to F or D, and C must not sit next to D. Put A-E-C on the right and the only way to keep A away from F and D is to read it as C-E-A, so A is at the far end. That makes C the left neighbour of the F-B-D block, and the block must then end in F, not D, to keep C away from D. The row becomes D-B-F-C-E-A. Checking: B is between D and F; E is between C and A; A touches only E; C touches F and E; D touches only B.

Reversing the blocks gives A-E-C-F-B-D, and again F's neighbours are B and C. So the answer is C, B and C, in every valid arrangement.

Core concept: linear seating or arrangement with blocked pairs, where you lock the fixed sub-groups first and then test the negative conditions at the joins. This is a staple of UPPSC reasoning and sets up the same method for Mains-style logic questions.

Trap: aspirants often place B-D-F without checking which end touches the other block, then lose track of the C-D restriction and guess between B and D as F's neighbour.