(GR. 10) A table has $9$ seats,$4$ on one side facing the main door and $5$ on the opposite side facing the garden. How many...

Question

So I managed to answer this question and I got $4,320$ possible ways $(5! \cdot 3!\cdot 3!)$. Unfortunately, the answer in my textbook is $43,200$ possible ways, but this wouldn’t be the first misprint my book had. If anyone would be willing to give me a shove towards the correct answer, I would really appreciate it!

“A table has $9$ seats, $4$ on one side facing the main door and $5$ on the opposite side facing the garden. In how many ways can $9$ people be seated at the table if Miguel, Eric, and Sam must sit on the side facing the garden?”

Answer 1

Assuming "the side facing the table" means "the side facing the garden":

First, we choose the seats for Miguel, Eric, and Sam. This can be done in $5\times 4\times 3=60$ ways.

Then, there are $6$ seats left. So $6!=720$ ways to arrange the remaining people.

Combining the two gives $60\times 720=43200$.

Answer 2

You could think as follows:

You put Eric, Miguel and Sam on the side with $5$ chairs and pick $2$ more people to fill this side.

You could choose these people in $\binom{6}{2}$ different ways and now these people will sit on the side with with $5$ chairs so you could arrange them in $5!$ different ways. This means that so far we have $\binom{6}{2}\cdot 5!$ different possibilities.

Now the rest of the $4$ people can be arranged in $4!$ different ways on the side with $4$ chairs. This means that we have

$$ \binom{6}{2}\cdot 5!\cdot 4!=43.200 $$ different ways to arrange them according to the given rules.