Twelve people travelling in three cars I have to solve an introductory counting principles problem, It goes like this:


Twelve people travel in three cars, with four people in each car. Each car is driven by its owner. Find the number of ways in which the remaining nine people may be allocated to the cars. (The arrangement within the car doesn't matter)


I thought that the nine people have to be grouped into 3 cars where the arrangement doesn't matter, so the answer should be:

9C3
  = 84 

But, this is not the right answer. So, how do i solve this?

Correct answer is ${9\choose3}\times{6\choose3}$.
$3!$ is not needed as it is already factored in ${9\choose3}\times{6\choose3}$. It's easy to understand if we try with small examples.
