Here when the right and left neighbours are interchanged, they have to be counted separately. Rotations of a given configuration are considered the same. These are criteria here based on which we have to calculate the permutations.
The easiest way would be to fix a given person and then we have $3! = 6$ ways of arranging the remaining positions. By doing this way, the repeated counting of rotations have been eliminated and left and right neighbours interchanged are counted separately.