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. $^5p_3$ = 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 therefore $3*(^4P_2)$
So ans should be $^5p_3$ - $3*(^4P_2)$ = 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 ??
 A: You ask for any other approach.
Since the numbers are small we can split the cases up according to the number of originally unoccupied chairs that are used.
No originally unoccupied chair used
Then the first person can move chair in $2$ ways and that fixes where everyone goes.
One used
We can choose the chair and the person who now sits in it in $2.3=6$ ways. The other two people either switch seats or one of them sits in the newly unoccupied seat. So each of the $6$ ways has $3$ arrangements.
Both used
There are $3.2$ arrangements in the two originally unoccupied chairs. The other person can now move chair in $2$ ways.
The total is $2+18+12=32$.
A: When the numbers are small, there is nothing wrong with forgoing all elegance, taking off your shoes, and counting on your fingers and toes.
Assume that Person-1 is originally seated in Seat-1, Person-2 in Seat-2, and Person-3 in Seat-3.
There are two basic cases:
Case 1: Person-1 shifts to Seat-2 (which was already occupied) 
Case 2: Person-1 shifts to Seat-4 (which was not already occupied).
Case 1: Person-1 shifts to Seat-2
You will end up multiplying the enumeration by $2$, to include the case of Person-1 shifting to Seat-3, which by symmetry, should have the same analysis.
Assuming that Person-1 shifts to Seat-2, if Person-2 shifts to Seat-3, then Seats 2 and 3 are now taken, so there are 3 seats that Person-3 can take. If Person-2 shifts to any of the other remaining seats (i.e. Seat-1, Seat-4, Seat-5), then there will only be two Seats that Person-3 can shift to.
Therefore, in Case-1, if Person-1 shifts to Seat-2, the total number of new seating arrangements is
$$(1 \times 3) + (3 \times 2) = 9.$$
As indicated at the start of Case-1 analysis, you need to multiply this by $2$, since by symmetry, you get the same analysis if Person-1 shifts to Seat-3.
Therefore, the total for Case-1 is
$$2 \times [(1 \times 3) + (3 \times 2)] = 18.$$
Case 2: Person-1 shifts to Seat-4
You will end up multiplying the enumeration by $2$, to include the case of Person-1 shifting to Seat-5, which by symmetry, should have the same analysis.
If Person-1 shifts to Seat-4, and Person-2 shifts to Seat-3, then there are 3 seats that Person-3 can shift to.  If Person-1 shifts to Seat-4 and Person-2 shifts to either Seat-1 or Seat-5, then there are only two seats that Person-3 can shift to.
Therefore, in Case-2, if Person-1 shifts to Seat-4, the number of possible new seating arrangements are
$$(1 \times 3) + (2 \times 2) = 7.$$
As in Case 1, this needs to be multiplied by $2$, since Person-1 could (similarly) shift to Seat-5.
Therefore, in Case-2, the enumeration is
$$2 \times [(1 \times 3) + (2 \times 2)] = 14.$$

Final Answer:
$$14 + 18 = 32.$$
