A simple question in combinatorics.

Question

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.

Answer 1

Let's see how you arrived at $3^6$.
Each seat can be occupied by $3$ different people, and there are $6$ seats, so the number of possibilities is $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.
So one of these $3^6$ possibilities is: First seat is occupied by Professor. Second seat has $3$ choices: Professor, Student, Clerk. Second seat selects Professor. Wait a minute, now the first and second seats are fighting for the professor. Professor can't occupy both seats!
So, no, $3^6$ is horribly wrong.

Is it $6^3$, then? Each person has $6$ choices of seats, and there are $3$ people, so $6 \times 6 \times 6 = 6^3$. Now we have the problem of the same seat being occupied by different people. So that's not it, either. But it can be fixed, by eliminating the selected seats. So, Professor selects one out of $6$ seats, then Student selects one out of $5$ seats, and Clerk selects one out of $4$ seats, so the answer is $6 \times 5 \times 4$ (exactly the same as when there are $6$ people and $3$ seats, because seats and people are not different in any relevant manner).

Another way is, you have three dummies to occupy three seats, but they are identical. So using the idea of permutation with repetition (see http://www.math.wsu.edu/students/aredford/documents/Notes_Perm.pdf, Page $6$), we have $\dfrac{6!}{3!} = 6 \times 5 \times 4$ possibilities.

Answer 2

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 $$ 3!\binom 63=3!\frac{6!}{\left(3!\right)^2}=\frac{6!}{3!}=6\cdot5\cdot4 $$