A classroom has two rows of eight seats each. There are 14 students, 5 of who always sit in the front row and 4 of who always sit in the back row. In how many ways can the students be seated?
$\dfrac{8!}{3!}\dfrac{8!}{4!}\dfrac{7!}{2!} = 28449792000$
The correct answer might be clearly obvious but I'm asking what is wrong with the approach below.
There are 14 students and 5 like to sit in the front row. How can we tell which five? I'd pick 5 out of 14 and permute them in the first row. Then I'd pick 4 out of 9 and permute them. Which leads to a different solution.
$\dbinom{14}{5}\dfrac{8!}{3!}\dbinom{9}{4}\dfrac{8!}{4!}\dfrac{7!}{2!} = 7176516931584000$
Now, what is wrong with the picking of the 5 and 4 kids that like to sit in the first/last row? And is there something in the wording of the problem where one can assume that kids are already picked and there's no need for applying the picking process?
Thanks!