Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is

We need to find the number of ways to transport 8 persons in 3 cars (of different makes), where each car can hold at most 3 persons.

Since each car holds at most 3 persons and the total is 8, the only valid partition is: 3 + 3 + 2 = 8. No other partition works (e.g., 4+3+1 would require a car with 4, exceeding the limit).

Since the three cars are distinguishable (different makes), we need to decide which car gets 2 persons and which two get 3 persons.

Method: First choose which car carries only 2 persons: there are $$\binom{3}{1} = 3$$ ways.

Then, choose 2 persons for that car: $$\binom{8}{2}$$ ways.

Then, choose 3 persons from the remaining 6 for the next car: $$\binom{6}{3}$$ ways.

The last 3 persons go to the remaining car: $$\binom{3}{3} = 1$$ way.

Alternative calculation: We can also compute this as $$\frac{8!}{3!3!2!} \times \frac{3!}{2!}$$, where $$\frac{8!}{3!3!2!} = 560$$ counts the ways to partition 8 into groups of (3,3,2), and $$\frac{3!}{2!} = 3$$ assigns distinguishable cars to these groups (dividing by $$2!$$ because the two groups of 3 are interchangeable in the partition, but then multiplied by $$3!$$ for assigning to distinct cars).

The correct answer is Option 3: 1680.