Twelve people travelling in three cars - clarification on existing answer Please refer this question 

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)

Selected answer of this question is ${9\choose3}\times{6\choose3}\times{3\choose3} \times3!$
But I believe this answer includes overcounting also. I think correct answer is ${9\choose3}\times{6\choose3}$ only (reference : explanation provided here for a similar question)
Am I correct here or my understanding is wrong and correct answer is still ${9\choose3}\times{6\choose3}\times{3\choose3} \times3!$?

Yes, you are right. 
The mistake I did was to take extra care of the dissimilarity of the cars, thus causing repetitions.  
Please check the updated answer, thanks :-)
