# A classroom has two rows of eight seats…

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!

-
I believe the intent of the question is that there are already five particular students who always sit in the first row. You do not have the freedom to choose which five they are. –  Austin Mohr Dec 29 '12 at 22:28
Yes, the five that prefer the front row are not changing with each "run." Assume you know their names, instead. –  Thomas Andrews Dec 29 '12 at 22:29
Is it, then, the word "particular" that one should look for? I'm asking because the same confusion appeared in a problem where one particular woman and one particular man shouldn't have been paired and I tried picking the woman and the man. –  Looft Dec 29 '12 at 22:31

Your mistake is in thinking that you get to pick which $5$ students always sit in the front row: you don’t. You have no choice in the matter: it’s a definite set of $5$ students who always sit in the front row. Similarly, it’s a definite set of $4$ students who always sit in the back row. Then there are $5$ students who sometimes sit in the front row and sometimes in the back.