Question: In how many ways can you seat 40 students on 10 different benches with 4 seats each? I was thinking $40 \* 39 \* 38 \* 37 + ... + 4 \* 3 \* 2 \* 1$, all this multiply by $10!$  
 Am I right? How should I write this in the shortest way if so?

Answer: 
The 10 rows with 4 seats each is a red herring. There are 40 seats and 40 students. It doesn't matter if the seats are in 10 rows of 4, 4 rows of 10, 3 rows of 7 and one row of 19 or whatever.

There are 40 seats for the first student, 39 for the second, etc. 

The answer is 40!.

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After some thought, I think a case could be made for other answers. If we imagine the rows as having imprinted letters A-J on the side, then maybe we can argue that placing Row A in front is different then placing row G in front. In which case there are 10! ways to arrange the rows and then 40! to place the students for 10!40!

I don't think we *should* interpret it that way. And if I wanted to nitpick I could argue that there are far more ways the 10! to arrange the benches; I could arrange them in a square facing each other, in rows behind each other, in a circle. Three of them can be inside the building and 7 of them outside. etc. And can we rearrange the seats within each row? Can we move seat number 39 from row J to row C?

But we *could* interpret it that way (the order of the benches matter). And it is legitimate. Why else would it specify that the benches are different or that the seats are in benches at all?

I think we are fast approaching a "Mr. Left" problem. When elementary children are given word problems they frequently look for "code words". "and", "together", "all" = plus, "left", "remaining", "away" = minus. So when given a problem "Mr. Left has two children aged 5 and 7? How old is the older one?" a surprising and depressing number will answer the question with "2". Why? Because "left" means to subtract and 7 -5 = 2, of course.

So here we are told "different" means "order matters". "circular table" means "the order of the first choice doesn't matter but the rest do". "replacement" means $n^m$ but "without replacement" means ${n \choose m}$, etc.

So... I don't know whether we are supposed to count the different ways the benches can be arranged. If we are, I do wish to argue that it wasn't clear to *me*. After all, so what if the benches are different? Of course they are different! We don't have a transdimensional-cohabitor-flux device where you can have 10 benches that are the same bench, do we?

Okay, rant over. 

Your guess is as good as mine. I don't care what the actual answer is as long as it can be validly justified and I can see at least four such different answers.
This depends on whether the order of students on each bench matters. If it does, then it's just $ 40! $, the 10 benches bit is a red herring. If it doesn't, then it's $ \binom{40}{4} \times \binom{36}{4} \times \cdots \times \binom{4}{4} $, i.e. number of ways to pick 4 for the first bench, multiplied by number of ways to pick 4 for the next bench (out of the 36 remaining students) etc.

This is assuming all benches are identical.