A, B, C, D, E and F are six HoDs of a college and

Solution

This is a circular arrangement puzzle. B is between C and D. So possible arrangements: ... C - B - D ... or ... D - B - C ... E is second to the left of D and second to the right of C. This means if we go clockwise, the order is C, ?, E, D. So E is between C and D with one person in between. Since B is between C and D, that one person must be B. So the order is C - B - E - D (clockwise). Now, A is second to the left of F. Only F is left to be placed. The clockwise order so far is C, B, E, D. The next two positions complete the circle. Place F and A. A is second to the left of F. So the order must be A, ?, F. The only way to fit this is to have the sequence: D, A, ?, F, C. But we have E between B and D. Let's list all points in order based on point 2 (C-B-E-D). The next after D is the empty seat, then the next is the last empty seat, then back to C. Let's assume clockwise order: C -> B -> E -> D -> X -> Y -> C. A is second to the left of F. So if F is at position X, then second to left is position E (but E is already occupied). If F is at position Y, then second to left is position D (occupied). So let's place A and F in X and Y. If F is at Y, then second to its left is X. So A must be at X. Then the order is: C, B, E, D, A, F.