A simple question in combinatorics. A university bus stops at some terminal where one professor,one student and one clerk has to ride on bus.There are six empty seats.How many possible combinations of seating?
My problem:I know that if there are six people to fill the 3 seats then there are $6\times5\times4=120= ^6P_3$$\hspace{0.3cm}$ possible combinations.
If there are 3 people to fill six seats then there will be $\hspace{0.3cm}$$3^6$$\hspace{0.3cm}$ possible combinations.Am i right?
so answer of my question should be$\hspace{0.3cm}$$3^6$.Am i right?
Thanks in advance.  

One way to solve it is to first place one person. This can be done in $6$ ways. Then the second ($5$ ways) and than the third ($4$ ways). This gives $6\cdot 5\cdot 4=P^6_3$ possibilities.
Another way (which I prefer) is to first choose $3$ of the $6$ seats for the three persons in $\binom 63$ ways. Then, we can distribute the three people in $3!$ ways over the three seats, resulting in
$$
3!\binom 63=3!\frac{6!}{\left(3!\right)^2}=\frac{6!}{3!}=6\cdot5\cdot4
$$
