Solving two questions on combinations I have recently learned permutations and combinations. While I was practising, I couldn't solve these two questions:

In how many ways can a group of 14 people eating at a restaurant be divided between 3 tables seating 6, 5 and 3?

This is what I did. So I did:
  14C6 + (14 - 6)C5 + 3C3
= 14C6 + 8C5 + 3C3
= 5369

But the answer was actually, 168168.
Also,

A group of 12 guests at a wedding are to travel as passengers from the church to the reception. There are 3 cars: black, silver and blue. Each car holds 4 passengers. Find the number of ways in which the group may travel if Alice and Jack refuse to travel in the same car?

So I calculated how many ways passengers could be arranged with Alice and Jack possibly meeting.
  12C4 + (12 - 4)C4 + 4C4
= 12C4 + 8C4 + 4C4
= 566

Then I wasn't sure what to do after.
Could someone tell me what I am doing wrong and how to continue on from the second question?
Thanks in advance.
 A: In the first one you should take the product and not the sum of those binomial coefficients (see the Rule of product)
$$\binom{14}{6}\binom{14-6}{5}\binom{14-6-5}{3}=\binom{14}{6}\binom{8}{5}\binom{3}{3}=168168.$$
As regards the second one this is my hint. We have $\binom{12-2}{3}$ ways to fill Alice's car, $\binom{12-2-3}{3}$ ways to fill Jack's car. The remaining $4$ guests go in the other car. Now assign the $3$ colors to the $3$ cars:  Alice's car can have $3$ colors, Jack's car $3-1$ colors... 
Can you take it from here?
A: For the second problem, the solution is as follows :
Assume Alice and Jack travel in car's as follows :
Assume Alice goes to car 1 and Jack goes to car 2. The number of ways to arrange the other people :
$\binom{10}{3} * \binom{10-3}{3} * \binom{10-3-3}{4} = 4200$.
Actually the number if ways is same for rest of the cases i.e. $(A2, J1), (A2, J3), (A3, J2), (A1, J3), (A3, J1)$. where $A_i = $ Alice goes to car $i$ and $J_i = $ Jack goes to car $i$.
Since, these events independently lead to final answer, we apply rule of addition and the final answer is 
$4200 + 4200 + 4200 + 4200 + 4200 + 4200 = 25200$.
