How many ways can 40 people be split into 10 quartets? "A certain music school has 49 students, with 10 each studying violin, viola, cello, and string bass. The director of the school wishes to divide the class into 10 string quartets; the four students in each quarter study the four different instruments. In how many ways can this be done?"
This is question from my discrete math textbook. It's under partitions and equivalence classes. I have no idea how to start this problem. Any hints?
I can't seem to justify my answers enough to convince myself. I think the answer that makes most sense to me is:
40! / (10!4!)
Because there are 40 possible students (no repetition allowed), and they are partitioned into 10 groups of 4-unique elements. 
 A: Line up the violinists say in alphabetical order.  
Now pick a violist, a cellist, and a bass for the first violinist ($10^3$ ways).
Now pick a violist, a cellist, and a bass for the second violinist ($9^3$ ways).
And so on.
Do you see what happens from here?
A: Alternatively, after lining up the violinists as in paw88789’s answer, you can choose all of the violists at once by lining them up in front of the violinists; how many ways are there to line up $10$ people? Then do the same with the cellists, and again with the bassists. It’s the same idea, but with the choices made one instrument at a time instead of one quartet at a time.
The key to both answers is fixing the order of the violinists: that’s what ensures that you don’t count any division into quartets twice.
Added: Your $\frac{40!}{10!4!}$ can be interpreted in the following way. There are $40!$ ways to line up all $40$ students. Say that we don’t care in which order the ten cellists appear in the lineup: keeping their overall positions in the lineup, they can be permuted in $10!$ different ways, so we divide by $10!$. The fraction $\frac{40!}{10!}$ then counts the number of ways to line up the $40$ students if we treat the cellists as indistinguishable, i.e., we care only where they are in the lineup, not which one is in which of those $10$ positions. Finally, we decide that we’ll also treat the four tallest violists as indistinguishable: we care which $4$ positions they occupy as a group, but we don’t care which of them is in which of those positions. That gives us $\frac{40!}{10!4!}$ lineups that differ in ways that we care about. I think that you can probably see that this bears little resemblance to what we want to count.
An alternative correct argument based on dividing out unwanted duplicates could go like this. Say we have quartets $1$ through $10$. There are $10!$ ways to assign the violinists to them, $10!$ ways to assign the violists to them, and so on, for a total of $10!^4$. However, the numbering of the quartets was arbitrary: if I permuted the numbers, which I can do in $10!$ ways, I’d still have the same quartets. To correct for that, I divide by $10!$, getting $10!^3$.
