Arranging 6 different people in all possible ways in a 3 floor house Say we have a house that fits 6 people. All the rooms on the left side of the house have big rooms, all the rooms on the right side have small rooms
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Floor 3| A | B |
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Floor 2| A | B |
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Floor 1| A | B |
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We have 6 people to throw in this house now.
Say Jeremiah gets $Floor\ 3\ A$ bedroom and Todd gets $Floor\ 3\ B$. If they were to switch rooms, this would be considered a different layout for them, solely because the rooms are not equal.
The floors are not equal either. If the exact arrangement above is duplicated on $Floor\ 2$ that is different than if they were in $A$ and $B$ respectively, on $Floor\ 3$.
So with 6 people in this house with possible orderings counting, I calculate there being $6!$ different ways for the house to be filled out.
Say we go to another house where the rooms are equal but the floors are not.
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Floor 3| = | = |
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Floor 2| = | = |
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Floor 1| = | = |
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Now we know if Jeremiah and Todd are on $Floor\ 3$, there can only exist one way they can be on that floor since the rooms are equal and  interchangeable. So it seems that since every floor's permutation counts as only one ordering, the answer for this house arrangement is $\frac{6!}{2!}$
Now say we go to another house where the rooms and floors are equal. This means if Jeremiah and Todd are together, in what rooms on what floors is entirely irrelevant. The only information we need to know is who is together as a pair and that is all. For this house arrangement I see $\frac{6!}{2!\times3!}$ different possibilities.
I am wondering if the way I am looking at narrowing down these orderings is correct or if I am missing something in my division or thought process. Thanks!


So it seems that since every floor's permutation counts as only one ordering, the answer for this house arrangement is $\frac{6!}{2!}$

This part of the answer is wrong. Since there are 3 floors, on each of which the permutations don't matter, the denominator will change as:
$$\frac{6!}{2!^3}$$
Similarly:

For this house arrangement I see $\frac{6!}{2!\times3!}$ different possibilities.

This will become:
$$\frac{6!}{2!^3*3!}$$
