1. Interpretation of the problem

We have:

• $8$ distinct persons,
• $3$ cars of different makes arrow the cars are distinct,
• each car can carry at most $3$ persons.

Since all $8$ persons must be transported using $3$ cars, and each car can take at most $3$, the only possible distribution of persons among the cars is:

$3,3,2$

because:

$8 = 3+3+2$

No other split into three positive parts with each part $\le 3$ is possible.

2. Count the assignments to distinct cars

Let the three distinct cars be $C_1, C_2, C_3$.

We must assign:

• $3$ persons to one car,
• $3$ persons to another car,
• $2$ persons to the remaining car.

Because the cars are distinct, first choose which car gets $2$ persons.

This can be done in:

$3 \text{ ways}$

3. Choose the persons for each car

Once the car carrying $2$ persons is fixed:

• Choose $2$ of the $8$ persons for that car: $\binom{8}{2}$

• From the remaining $6$ persons, choose $3$ for one of the two remaining cars: $\binom{6}{3}$

• The last $3$ automatically go to the last car: $\binom{3}{3}=1$

So total ways are:

$3 \cdot \binom{8}{2} \cdot \binom{6}{3}$

Now compute:

$\binom{8}{2} = 28, \qquad \binom{6}{3} = 20$

Hence,

$3 \cdot 28 \cdot 20 = 1680$

4. Check against options

$1680$

So the correct option is:

B

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.