TITLE: Why does this intuition for combinations with repetition fail?
QUESTION [7 upvotes]: When you are learning the difference between combinations and permutations without repetition. The logic for each of them is:
1) Permutation without repetition: Selecting 4 objects from 10, gives $10*9*8*7$ choices.
2) Combinations without repetition: Selecting 4 objects from 10, gives $10*9*8*7/(4*3*2*1)$.
Here you always divide by the number of times you can arrange 4 objects since each of those combinations are the same, order doesn't matter.
3) Permutations with repetition: selecting 4 objects from 10, is $10*10*10*10$
4) Combinations with repetition
Now trying to apply this to combinations with repetition, my reasoning is this. Its similar to how one goes from permutations to combinations in the case of no repetition.
There are 10 objects to choose from and you want to select 4. Each selection doesn't diminish the number of choices you have since its replacable. So for each possible selection you have 10 choices, selecting 4 then gives $10^4 / 4!$ since you need to divide by the number of ways 4 objects can be arranged to not overcount. 
I know this is wrong, I've seen solutions to this using stars and bars. But my question is basically, for the case of no repetition, combinations can be treated as permutations only if you divide by the number of ways you can arrange your selection, and that makes intuitive sense to me.
When it comes to repetitions, permutations make sense to me. Selecting k objects from n choices is just $n^k$. So why cant I use the same reasoning here and say that the number of combinations is $n^k / k!$?

REPLY [3 votes]: There are 24 variants of: ABCD, ABDC, ... , DCBA. These are all distinct permutations without repetition, and they all condense to the same combination without repetition. This accounts for the division by $4!$ when there is no repetition.
But consider these permutations with repetition: AABB, ABAB, ABBA, BAAB, BABA, BBAA. For two As and two Bs, that's it. Now six permutations condense to one combination.
And here: AAAB, AABA, ABAA, BAAA. We have four permutations condensing to one combination.
And here: AAAA. Just one permutation condensing to one combination.
And here: ABCC, ACBC, ACCB, BACC, BCAC, BCCA, CABC, CACB, CBAC, CBCA, CCAB, CCBA. We have twelve condensing to one.
So there is no consistent ratio. Depending upon the makeup of the combination, there are somewhere between $1$ and $24$ permutations. It varies over all the combinations with repetition.