Permutations and Combinations: Mathematics | JEE Main

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:

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A3360B1680C560D1120

Step-by-step solutionView

Correct answer

Partition 8 people into group sizes of 3, 3, and 2, then distribute these groups across the 3 distinct cars, giving a total of 1680 ways.

Option analysis

Why each option works or fails

A · 3360

Forgetting that swapping the two groups of size 3 among themselves does not create a new partition before assigning to cars, leading to double-counting. When dividing into groups of sizes 3, 3, and 2, divide by 2! to account for the two identical group sizes before multiplying by 3! to assign them to distinct cars.

B · 1680

This is the correct option. Partitioning 8 people into sizes (3, 3, 2) gives 8! / (3! 3! 2! 2!) = 280 ways, and assigning them to 3 distinct cars gives 280 × 3! = 1680 ways.

C · 560

Computing the number of ways to partition the 8 people into sizes (3, 3, 2) without assigning each group to a specific car. Because the cars are of different makes, you must multiply the number of group partitions by 3! / 2! or select which car receives 2 people (3 ways), giving 560 × 3 = 1680.

D · 1120

Making an arithmetic error in evaluating the combinations, such as undercounting the car permutations. Multiply the combination choice 8C2 × 6C3 × 3C3 = 28 × 20 = 560 by 3 (the number of choices for which car gets 2 people) to get 1680.

Reviewed route

Solution

StepWorking

01given

Number of persons $n = 8$ , number of distinct cars $= 3$ , and maximum capacity of each car is $3$ persons.

02goal

Find the total number of ways to transport all 8 persons in the 3 distinct cars such that no car has more than 3 persons.

03approach

Find all integer partitions of 8 into 3 parts where each part is $\le 3$ . Since $3+3+2 = 8$ , the only valid partition of group sizes is $(3, 3, 2)$ . We then partition the 8 distinct people into groups of sizes 3, 3, 2, and assign these 3 groups to the 3 distinct cars.

04execute

Number of ways to divide 8 distinct persons into groups of sizes 3, 3, 2 is $\frac{8!}{(3!)^2 \cdot 2! \cdot 2!}$ . Then, assign these groups to the 3 distinct cars by multiplying by $3!$ : $\text{Ways} = \frac{8!}{3! 3! 2! 2!} \times 3! = \frac{8!}{3! 2! 2!} = \frac{40320}{6 \times 2 \times 2} = 1680$

✓verify

Alternative selection method: Pick 3 people for Car 1: $\binom{8}{3} = 56$ . From remaining 5, pick 3 for Car 2: $\binom{5}{3} = 10$ , and remaining 2 go to Car 3: $\binom{2}{2} = 1$ . Multiply by the 3 arrangements of car sizes $(3, 3, 2)$ : $56 \times 10 \times 1 \times \frac{3!}{2!} = 560 \times 3 = 1680$ . Matches correctly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why do we divide by

2!

in the denominator of the grouping formula?

There are two groups of the same size (3 and 3). Partitioning 8 distinct people into unlabeled groups of sizes 3, 3, 2 requires dividing by $2!$ . This avoids overcounting identical grouping combinations.

Answer

Partition 8 people into group sizes of 3, 3, and 2, then distribute these groups across the 3 distinct cars, giving a total of 1680 ways.

Why each option works or fails

A: 3360 - Forgetting that swapping the two groups of size 3 among themselves does not create a new partition before assigning to cars, leading to double-counting. When dividing into groups of sizes 3, 3, and 2, divide by 2! to account for the two identical group sizes before multiplying by 3! to assign them to distinct cars.
B · correct: 1680 - This is the correct option. Partitioning 8 people into sizes (3, 3, 2) gives 8! / (3! 3! 2! 2!) = 280 ways, and assigning them to 3 distinct cars gives 280 × 3! = 1680 ways.
C: 560 - Computing the number of ways to partition the 8 people into sizes (3, 3, 2) without assigning each group to a specific car. Because the cars are of different makes, you must multiply the number of group partitions by 3! / 2! or select which car receives 2 people (3 ways), giving 560 × 3 = 1680.
D: 1120 - Making an arithmetic error in evaluating the combinations, such as undercounting the car permutations. Multiply the combination choice 8C2 × 6C3 × 3C3 = 28 × 20 = 560 by 3 (the number of choices for which car gets 2 people) to get 1680.

Step-by-step solution

given: Number of persons $n = 8$ , number of distinct cars $= 3$ , and maximum capacity of each car is $3$ persons.
goal: Find the total number of ways to transport all 8 persons in the 3 distinct cars such that no car has more than 3 persons.
approach: Find all integer partitions of 8 into 3 parts where each part is $\le 3$ . Since $3+3+2 = 8$ , the only valid partition of group sizes is $(3, 3, 2)$ . We then partition the 8 distinct people into groups of sizes 3, 3, 2, and assign these 3 groups to the 3 distinct cars.
execute: Number of ways to divide 8 distinct persons into groups of sizes 3, 3, 2 is $\frac{8!}{(3!)^2 \cdot 2! \cdot 2!}$ . Then, assign these groups to the 3 distinct cars by multiplying by $3!$ : $\text{Ways} = \frac{8!}{3! 3! 2! 2!} \times 3! = \frac{8!}{3! 2! 2!} = \frac{40320}{6 \times 2 \times 2} = 1680$
verify: Alternative selection method: Pick 3 people for Car 1: $\binom{8}{3} = 56$ . From remaining 5, pick 3 for Car 2: $\binom{5}{3} = 10$ , and remaining 2 go to Car 3: $\binom{2}{2} = 1$ . Multiply by the 3 arrangements of car sizes $(3, 3, 2)$ : $56 \times 10 \times 1 \times \frac{3!}{2!} = 560 \times 3 = 1680$ . Matches correctly.