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 $n^k / k!$?