JEE Mathematics 2023 Question with Solution

Given: There are $8$ persons and $3$ cars of different makes. Each car can accommodate at most $3$ persons.

Find: The number of ways to transport all $8$ persons.

Since each car can hold at most $3$ persons and there are $8$ persons in total, the only possible distribution is $3,3,2$.

Choose $3$ persons for the first car:

$$\binom{8}{3} = 56$$

Choose $3$ of the remaining $5$ persons for the second car:

$$\binom{5}{3} = 10$$

The remaining $2$ persons go to the third car:

$$\binom{2}{2} = 1$$

Therefore, the number of assignments to three distinct cars is

$$\binom{8}{3}\binom{5}{3}\binom{2}{2} = 56 \times 10 \times 1 = 560$$

Because the cars are of different makes, the group sizes $3,3,2$ can be assigned to the three cars in

$$3! = 6$$

ways.

Hence, the total number of ways is

$$560 \times 3! = 560 \times 6 = 3360$$

So the solution concludes the correct option is C, which corresponds to $3360$. The answer key lists $1680$, but that disagrees with the extracted solution working.

Given: The $8$ persons must be split among $3$ distinct cars, with maximum capacity $3$ per car.

Find: Total number of valid distributions.

The only feasible occupancy pattern is $3,3,2$.

First split the $8$ persons into groups of sizes $3,3,2$:

$$\frac{8!}{3!\,3!\,2!} = 560$$

Now assign these three groups to the three cars. Since two groups have the same size $3$, the extracted solution multiplies by $3!$ for assigning the groups to the cars:

$$560 \times 3! = 3360$$

Therefore, according to the provided solution, the correct option is C.