combinations (choosing objects from "same set".. product rule)

Concerning this question (well, this is just the simplest
representative example of the type of question I am having some
struggles with):

"There are 12 students in a class. In how many ways can the 12 students
take 3 different test if four students are to take each test?"

What first comes to mind is: C(12,4) * C(8,4) * C(4,4)

Choosing elements 4 at a time from what's left in the set.

But the conflict arises (in my brain) when I remember the other type of
example I was having trouble with (see
http://groups.google.ca/group/sci.stat.math/browse_thread/thread/a9d5ae1ae28f270b/0d50eae8ecb694a4?lnk=gst&q=Gaya&rnum=4#0d50eae8ecb694a4
)

The problem with my solution there was that I was basically dealing
with combinations (unordered selections) from sets that were not
disjoint (the set of 4 aces was not disjoint from the set of "51 other
things").

But back to the above example, aren't we breaking a rule here too? The
first selection takes from the whole set of 12 students, the second
takes from a subset of this, and the third takes from a subset of both.
How does this not break any rules (especially the product rule,which
we use implicity when multiplying the 3 terms)?
How is this different from the aces problem I posted earlier?

Most questions I see that deal with combinations/selections have a
large set, with say, N elements, and the solution involves selections
from disjoint subsets of this large set.

For example, set of 10 students, 4 are married, 4 are single, 2 are
engaged. How many ways to pick a group of 3 married, 2 single and 1
engaged from this set of 10 students?

Easy: C(4,3) * C(4,2) * C(2,1)

This makes perfect sense as we are treating the set of 10 students as 3
separate subsets (disjoint) and making selections on them
independently. There is no way the product rule can go wrong here.

Even the corrected solution to the "at least n aces" problem goes
through the trouble of segregating the set of cards into 2 clearly
disjoint sets - this seems like an important way to keep things in
line.

When we make selections from non-disjoint sets (like the "at least n
aces" problem I linked above), we run into problems. Yet for some
questions it's valid to choose from the same set in successive C(n,r)
terms [like the 3 tests problem I've shown above].

What makes it "okay" to use the product rule on combinations/selections
from non-disjoint sets in this case, vs. "not okay" to use the product
rule on comb/sel. in the "at least n aces" case?

Any help will be greatly appreciate. Thank you to all who responded to
my "at least n aces" post earlier. I did not post a thank you reply
there because a discussion had started about something unrelated (FTC).

.

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