As promised in my response to Byron, here is my very partial progress to
date on this problem. Though long-winded, it doesn't amount to much.
I'm still looking for an answer to the general case, or to other special
cases (like $n=2$ or $c=2$) or even just better estimates. I'm also
interested to know about any applicable combinatorial tools even without
a full answer.
First, a caution: The graphs I'm about to use are not related to the
graphs in the general formulation of the original question. Ignore
those graphs and just think of students and classrooms!
Let $G$ be a graph with a vertex for each student, $cn$ vertices in all.
Say that a given year's assignment of students to classrooms
respects $G$ if every edge in $G$ joins two students that are
assigned to different classrooms. That is, $G$ encodes constraints on
assignments; each edge in $G$ represents two students that must be kept
apart. Let $R(G)$ be the probability that a random assignment of
students to classrooms respects $G$. Note that $R(G)$ does not depend
on $y$ since the definition of "respects" refers to a single year's
assignment.
Now let $e(G)$ be the number of edges in $G$. Then the probability that
after $y$ years every pair of students has at some point shared a
classroom is exactly
$$\sum (-1)^{e(G)} (R(G))^y$$
where the sum is taken over all graphs $G$.
This result follows from a straightforward application of
Inclusion/Exclusion. It's very pretty (at least I think so) but useless
if you want a real answer. The problem isn't finding $R(G)$; this is tedious
but can be automated. The real difficulty is that you have to handle
$2^{cn\choose 2}$ graphs, which is far too many. The graphs come in
isomorphic bunches (e.g. there are $cn\choose 2$ one-edged graphs that all
have the probability of being respected) and you don't have to handle most graphs with lots of edges since $R(G) = 0$ for any graph $G$ that is not $c$-partite. But this doesn't help enough.
We get some use out of this wretched formula by generalizing the
problem. Suppose we ask, given the same conditions, for the probability
that each student in a given group of $p$ students at some point sees
each of the other students in that group. The formula above gives the
exact answer if we sum over all graphs with $p$ vertices, each labelled
with a student in the group. For example, with $p = 2$ the formula
gives the probability that two given students at some point see each
other:
$$1 - \left(\frac{n(c-1)}{cn-1}\right)^y$$
(Of course we can get this result by a simpler route.) For $p=3$ the
formula gives
$$1 - \left(\frac{N-n+1}{N(N-1)}\right)^y(3(N-1)^y - 3(N-n)^y + (N-2n+1)^y)$$
where $N = cn-1$. I've fought through the formula for $p$ up to 6, but
it becomes impossible quickly.
Now some approximations. Define $U_m$ to be the probability that a
given student $S$ is at some point in a classroom with each student in a
given set of $m$ students not containing $S$. Another straightforward
application of inclusion/exclusion gives
$$U_m = \sum_{j=0}^m(-1)^j{m\choose j}{N-j\choose n-1}^y \bigg/ {N\choose n-1}^y$$
where $N = cn-1$ as before. $U_N$ is then the probability that a single
given student will see all the others. This is a very weak upper bound
on the probability that all the students see all the others, though in
trivial cases one student seeing all the others does imply that they all
do! $U_m$ is rather easy to calculate; with the original parameters,
for example, $U_{74}$ turns out to be a bit less than 1/7792. Note, BTW,
that $U_1$ correctly gives the same result as the $p=2$ answer above.
If "student $A$ sees all the others" were independent of "student $B$
sees all the others", then we'd have the answer we seek: $(U_N)^{cn}$.
But of course these events are unlikely to be independent. If many
students have seen all the others, then surely the probability increases
that all have seen the others. So $(U_N)^{cn}$ may be a lower bound on
the true answer, if a weak one. Edit: Actually, I don't think it can be a lower bound, since I'm pretty sure one could find a case where one student can see all the others, but it's impossible for all students to do so.
Here's another way of estimating the answer: A group of $p$ students
all see each other if and only if both (a) a particular one of these
students sees all the other $p-1$ students, and (b) the other $p-1$
students all see each other. The probability of (a) is exactly
$U_{p-1}$. Let $C_{p-1}$ denote the probability of (b). If events (a)
and (b) were independent, then the exact answer to the original question
would be $U_{p-1}C_{p-1}$. But by identical reasoning, $C_{p-1} =
U_{p-2}C_{p-2}$. Continuing in this way, the exact answer to the original
question would be
$$U_NU_{N-1}U_{N-2}\ldots U_2U_1$$
As above, the events in question are not likely to be independent. But
most of the time they're close, right? With the original parameters,
the fact that a group of five students all see each other has very
little impact on whether a sixth student sees each of the five, though
it should increase that probability slightly. This would make the
expression above a lower bound on the actual answer, though I don't know
how good it is.
