Question

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:

• A. 1120
• B. 560
• C. 1680
• D. 3360

Answer 1

The Correct Option is C

Step 1: Understanding the problem There are 8 persons to be transported, and there are 3 cars. Each car can carry at most 3 persons. We need to calculate the number of ways to assign these 8 persons to the cars. 
Step 2: Distribute the persons in the cars To distribute the 8 persons into 3 cars, with each car holding a maximum of 3 persons, we first assign 3 persons to two cars and 2 persons to the third car. The number of ways to select 3 persons for the first car from the 8 is given by: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] Now, 5 persons remain, so the number of ways to select 3 persons for the second car is: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \] Finally, 2 persons remain, so the number of ways to assign them to the third car is: \[ \binom{2}{2} = 1 \] 
Step 3: Consider the arrangements of the cars Since there are 3 cars of different makes, the arrangement of the cars is important. Therefore, the number of ways to assign the selected persons to the cars is multiplied by the number of ways to arrange the cars, which is 3! (since there are 3 cars). \[ \text{Total ways} = \frac{8!}{3!3!2!} = \frac{8 \times 7 \times 6 \times 5 \times 4}{4 \times 6} = 56 \times 30 = 1680 \] Thus, the number of ways in which the persons can be transported is 1680.