I encountered a very interesting problem from Black Book.
14 persons depart from a place in 2 cars of capacity 4 each and 2 auto - rickshaws of capaity 3 each. Find the number of ways in which they can depart if two particular persons insist on going in the same car (internal arrangement is not considered).
The cars and rickshaws are all distinguished.
This seems to be a Grouping and Distribution question.
My attempt:
Segregate the 2 people who insist on going together, and distribute the remaining 12 people in the 2 cars and 2 autos.
Of these 12 people, we know that 3 of them occupy each of the 2 autos, 4 of them occupy 1 car, and 2 of them occupy the other car. Additionally, 2! is added in the denominator to account for the ordering of the autos, to prevent overcounting in the selection.
So, the number of ways to distribute them is:
$$\dfrac{^{12}C_2 \times ^{10}C_3 \times ^{7}C_3 \times \times ^{4}C_4}{2!}$$
Simplifying gives
$$\dfrac{12!}{(3!)^{2}\times 4!\times 2!\times 2!}$$
After this, we account for the 2 people who insisted on going together, by settling them into the car having 2 people. There are $^2C_1$ ways of choosing 1 of the 2 cars having 2 free seats.
Hence, the expression becomes $$\cfrac{12!}{(3!)^{2}\times 4!\times 2!\times 2!}\times ^{2}C_1$$
But the answer given is
$$\cfrac{12!}{(3!)^{2}\times 4!\times 2!\times 2!}\times ^{2}C_1\times 2!$$
Here is my question: Where did the additional 2! come from? Is it because both the auto - rickshaws are identical in the number of people seated? Some input on the matter would be greatly appreciated.
I find your attempt and the book's difficult to fathom, and suggest that after putting the two "special" people in one of the two cars (in $2$ ways), just form labeled teams of $\,2,4,3,3\,$ people.
thus $\;2\times\Large\frac{12!}{2!4!3!3!}$