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Suppose you have 12 people, 4 of them are drivers. You have three cars, with one having 6 seats, the other 4 and the final 2 seats. In how many different ways can people be seated?

My solution was to count first the number of ways to seat the drivers, thus I concluded there would be $\binom{4}{3}$ to do so. Finally, all is left is to count the number of ways to seat the remaining people in the remaining seats, which would give me $\frac{9!}{5!3!2!}$. I conclude there are $$\binom{4}{3}\frac{9!}{5!3!2!}$$ ways to seat all the 12 people.

Yet, according to the solution, there are $$4!3!2! \frac{9!}{5!3!2!}$$ ways of doing it. My question is why is there $4!3!2!$ to seat the drivers and not $\binom{4}{3}$ ways?

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    Presumably, it matters which car is driven by which driver. – 2017-02-12
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    You have $\binom{4}{3}$ different groups of drivers, and then each of these groups have $3!$ possible car orderings (driver 1 can go to car 1, or to car 2 or to car 3, driver 2 idem, etc...) – 2017-02-12
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    Is it possible the solution is $$4\cdot 3!2! \frac{9!}{5!3!2!}$$ without the factorial on 4? – 2017-02-12

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The question isn't crystal clear, but taking a hint from the answer, I take it that which seat a person takes in a particular car doesn't matter, which driver gets which car does matter.

On the above assumptions, neither your answer nor the book's answer is correct.

Drivers can be assigned in $4\cdot3\cdot 2 = 4!$ ways,
and the remaining $9$ can be assigned in $\dfrac{9!}{5!3!1!}$ ways,

yielding an answer of $\;\;4!\cdot\dfrac{9!}{5!3!1!}$