I have a list of repeated items, such as

`list = {a, a, b, c, c, c}`

and I would like a list of *unique* Possibilities to choose 3 elements from it:

```
{{a, a, b}, {a, a, c}, {a, b, c}, {a, c, c}, {b, c, c}, {c, c, c}}
```

Unfortunately, "unique" in this sentence means two different things at the same time, and I cannot figure out how to achieve both types of uniqueness at the same time.

i could use `Permutations`

whose documentation regarding the input indicates that

Repeated elements are treated as identical.

But I'm going to have a lot of results that differ only by rearrangement, and I don't care about the order:

`Permutations[list, {3}]`

```
{{a, a, b}, {a, a, c}, {a, b, a}, {a, b, c}, {a, c, a}, {a, c, b}, {a, c, c}, {b, a, a}, {b, a, c}, {b, c, a}, {b, c, c}, {c, a, a}, {c, a, b}, {c, a, c}, {c, b, a}, {c, b, c}, {c, c, a}, {c, c, b}, {c, c, c}}
```

To eliminate the rearrangements, I could try using `Subsets`

but instead by *it is* Documentation,

Different occurrences of the same element are treated as different.

As a result, I get a lot of duplicate results that I don't want because of the repeated elements of `list`

::

`Subsets[list, {3}]`

```
{{a, a, b}, {a, a, c}, {a, a, c}, {a, a, c}, {a, b, c}, {a, b, c}, {a, b, c}, {a, c, c}, {a, c, c}, {a, c, c}, {a, b, c}, {a, b, c}, {a, b, c}, {a, c, c}, {a, c, c}, {a, c, c}, {b, c, c}, {b, c, c}, {b, c, c}, {c, c, c}}
```

[Aside from frustration, I can't imagine why Mathematica's permutation generation function treats repeated list items differently than the combination generation function.]

I could remove the duplicates from both results, but in both cases the full list of ambiguous results still needs to be calculated as an intermediate step, which I expect to be many orders of magnitude longer than the unique results.

Is it possible to get the result you want without first having to make an enormously longer list to get there?