Permutation / combination

Let car A , B , C hold 2 , 4 , 5 passengers respectively.
Assuming NOT consider the seat order in a certain car , otherwise , using permutation.Assuming these 9 people are NOT driver.
A1 , B3 , C5 :
9C1 * 8C3 * 7C5 = 9 * 56 * 21 = 10584 ways.
A1 , B4 , C4 :
9C1 * (8C4 * 4C4 / 2!) = 9 * (70 * 1 / 2) = 315 ways.
A2 , B2 , C5 :
(9C2 * 7C2 / 2!) * 5C5 = (36 * 21 / 2) * 1 = 378 ways
A2 , B3 , C4 :
9C2 * 7C3 * 4C4 = 36 * 35 * 1 = 1260 ways
A2 , B4 , C3 :
9C2 * 7C4 * 3C3 = 36 * 35 * 1 = 1260 ways
Total 10584 + 315 + 378 + 1260 + 1260 = 13797 ways.


2012-10-05 23:25:03 補充：
Corrections :

A1 , B3 , C5 :
should be 9C1 * 8C3 * 5C5 = 9 * 56 * 1 = 504 ways.

Total 504 + 315 + 378 + 1260 + 1260 = 3717 ways.

2012-10-06 11:56:09 補充：
Sorry for my mistakes , it is no need to divide 2! since the status of A and B are not the same.

Do it again :

A1 , B3 , C5 :
9C1 * 8C3 * 5C5 = 9 * 56 * 1 = 504 ways.

A1 , B4 , C4 :
9C1 * 8C4 * 4C4 = 9 * 70 * 1 = 630 ways.

A2 , B2 , C5 :
9C2 * 7C2 * 5C5 = 36 * 21 * 1 = 756 ways

2012-10-06 11:56:18 補充：
A2 , B3 , C4 :
9C2 * 7C3 * 4C4 = 36 * 35 * 1 = 1260 ways

A2 , B4 , C3 :
9C2 * 7C4 * 3C3 = 36 * 35 * 1 = 1260 ways

Total 504 + 630 + 756 + 1260 + 1260 = 4410 ways.

Thanks for remind.

2012-10-06 11:58:09 補充：
Please see Q3 of qid=7011011700773

2012-10-06 12:10:00 補充：
Alternatively :

A1 , B3 , C5 :
9! / (1! 3! 5!) = 504 ways

A1 , B4 , C4 :
9! / (1! 4! 4!) = 630 ways.

A2 , B2 , C5 :
9! / (2! 2! 5!) = 756 ways

A2 , B3 , C4 :
9! / (2! 3! 4!) = 1260 ways

A2 , B4 , C3 :
9! / (2! 4! 3!) = 1260 ways

Total 504 + 630 + 756 + 1260 + 1260 = 4410 ways.