Problem Statement

The set of combinations with repetitions is computed from a set,

S

{\displaystyle S}

(of cardinality

n

{\displaystyle n}

), and a size of resulting selection,

k

{\displaystyle k}

, by reporting the sets of cardinality

k

{\displaystyle k}

where each member of those sets is chosen from

S

{\displaystyle S}

. In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter. For example: Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets. Also note that doughnut can also be spelled donut.

Let's start with the solution:

var Combrep = Fn.new { |n, lst|
    if (n == 0 ) return [[]]
    if (lst.count == 0) return []
    var r = Combrep.call(n, lst[1..-1])
    for (x in Combrep.call(n-1, lst)) {
        var y = x.toList
        y.add(lst[0])
        r.add(y)
    }
    return r
}

System.print(Combrep.call(2, ["iced", "jam", "plain"]))
System.print(Combrep.call(3, (1..10).toList).count)


import "./seq" for Lst
import "./perm" for Comb

var a = ["iced", "jam", "plain"]
a = Lst.flatten(Lst.zip(a, a))
System.print(Comb.listDistinct(a, 2))

a = (1..10).toList
a = Lst.flatten(Lst.zip(Lst.zip(a, a), a))
System.print(Comb.listDistinct(a, 3).count)