I have a list with repeating elements, like

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

and I would like a list of *only* ways to choose 3 elements:

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

Unfortunately, "unique" means two different things at once in that sentence, and I can't figure out how to achieve both types of uniqueness simultaneously.

Could use `Permutations`

, whose documentation indicates regarding the entry that

Repeated elements are treated as identical.

But I will have many results that differ only by reorganization, 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 remove the rearrangements, you could try using `Subsets`

instead, but for *their* documentation,

Different occurrences of the same item are treated as different.

As a result, I get a lot of duplicate results that I don't want due to 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}}
```

[Frustrated aside: I can't begin to imagine why Mathematica's permutations-generating function treats repeating list items differently than its combinations-generating function.]

You could remove duplicates from any of the results, but either way, that still requires calculating the entire list of non-unique results as an intermediate step, which I hope will be much longer than the unique results.

Is it possible to get the result I'm looking for without first having to select a vastly longer list to get there?