Three people are seated in 3 our of 5 chairs . After a break , in how many ways can they be seated so that no person occupies the same same chair

I tried doing it by first counting the total arrangements i.e. 5P3 = 60 then subtracting the non desired arrangements fixing the first person at its place and arranging the other 2 person .This can done for the rest of the 2 people therefore3× 4P2
So ans should be 5P3 - 3(4P2) = 24
I know i have subtracted some same cases twice but i am not sure how to identify those cases.
Is there any other approach also ??