Listing combinations WITH repetitions in Scala

Question

Trying to learn a bit of Scala and ran into this problem. I found a solution for all combinations without repetions here and I somewhat understand the idea behind i

Answer 1

Daniel -- I'm not sure what Alex meant by duplicates, it may be that the following provides a more appropriate answer:

def perm[T](n: Int, l: List[T]): List[List[T]] =
  n match {
    case 0 => List(List())
    case _ => for(el <- l.removeDuplicates;
                sl <- perm(n-1, l.slice(0, l.findIndexOf {_ == el}) ++ l.slice(1 + l.findIndexOf {_ == el}, l.size)))
            yield el :: sl
}

Run as

perm(2, List(1,2,2,2,1)) 

this gives:

List(List(2, 2), List(2, 1), List(1, 2), List(1, 1))

as opposed to:

List(
  List(1, 2), List(1, 2), List(1, 2), List(2, 1), 
  List(2, 1), List(2, 1), List(2, 1), List(2, 1), 
  List(2, 1), List(1, 2), List(1, 2), List(1, 2)
)

The nastiness inside the nested perm call is removing a single 'el' from the list, I imagine there's a nicer way to do that but I can't think of one.