Combinations with repetitions

⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

The set of combinations with repetitions is computed from a set, $S$ (of cardinality $n$ ), and a size of resulting selection, $k$ , by reporting the sets of cardinality $k$ where each member of those sets is chosen from $S$ . 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: :Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e., $S$ is ${\mathrm{iced}, \mathrm{jam}, \mathrm{plain}}$ , $|S| = 3$ , and $k = 2$ .)

:A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.

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''.

;Task:

Write a function/program/routine/.. to generate all the combinations with repetitions of $n$ types of things taken $k$ at a time and use it to ''show'' an answer to the doughnut example above.
For extra credit, use the function to compute and show ''just the number of ways'' of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.

;References:

[[wp:Combination|k-combination with repetitions]]

;See also: {{Template:Combinations and permutations}}

Ada

Should work for any discrete type: integer, modular, or enumeration.

combinations.adb:

with Ada.Text_IO;
procedure Combinations is

   generic
      type Set is (<>);
   function Combinations
     (Count  : Positive;
      Output : Boolean := False)
      return   Natural;

   function Combinations
     (Count  : Positive;
      Output : Boolean := False)
      return   Natural
   is
      package Set_IO is new Ada.Text_IO.Enumeration_IO (Set);
      type Set_Array is array (Positive range <>) of Set;
      Empty_Array : Set_Array (1 .. 0);
      function Recurse_Combinations
        (Number : Positive;
         First  : Set;
         Prefix : Set_Array)
         return   Natural
      is
         Combination_Count : Natural := 0;
      begin
         for Next in First .. Set'Last loop
            if Number = 1 then
               Combination_Count := Combination_Count + 1;
               if Output then
                  for Element in Prefix'Range loop
                     Set_IO.Put (Prefix (Element));
                     Ada.Text_IO.Put ('+');
                  end loop;
                  Set_IO.Put (Next);
                  Ada.Text_IO.New_Line;
               end if;
            else
               Combination_Count := Combination_Count +
                                    Recurse_Combinations
                                       (Number - 1,
                                        Next,
                                        Prefix & (1 => Next));
            end if;
         end loop;
         return Combination_Count;
      end Recurse_Combinations;
   begin
      return Recurse_Combinations (Count, Set'First, Empty_Array);
   end Combinations;

   type Donuts is (Iced, Jam, Plain);
   function Donut_Combinations is new Combinations (Donuts);

   subtype Ten is Positive range 1 .. 10;
   function Ten_Combinations is new Combinations (Ten);

   Donut_Count : constant Natural :=
      Donut_Combinations (Count => 2, Output => True);
   Ten_Count   : constant Natural := Ten_Combinations (Count => 3);
begin
   Ada.Text_IO.Put_Line ("Total Donuts:" & Natural'Image (Donut_Count));
   Ada.Text_IO.Put_Line ("Total Tens:" & Natural'Image (Ten_Count));
end Combinations;

ICED+ICED
ICED+JAM
ICED+PLAIN
JAM+JAM
JAM+PLAIN
PLAIN+PLAIN
Total Donuts: 6
Total Tens: 220

AppleScript

-- combsWithRep :: Int -> [a] -> [kTuple a]
on combsWithRep(k, xs)
    -- A list of lists, representing
    -- sets of cardinality k, with
    -- members drawn from xs.

    script combsBySize
        script f
            on |λ|(a, x)
                script prefix
                    on |λ|(z)
                        {x} & z
                    end |λ|
                end script

                script go
                    on |λ|(ys, xs)
                        xs & map(prefix, ys)
                    end |λ|
                end script
                scanl1(go, a)
            end |λ|
        end script

        on |λ|(xs)
            foldl(f, {{{}}} & take(k, |repeat|({})), xs)
        end |λ|
    end script

    |Just| of |index|(|λ|(xs) of combsBySize, 1 + k)
end combsWithRep


-- TEST ---------------------------------------------------
on run
    {length of combsWithRep(3, enumFromTo(0, 9)), ¬
        combsWithRep(2, {"iced", "jam", "plain"})}
end run


-- GENERIC ------------------------------------------------

-- Just :: a -> Maybe a
on Just(x)
    {type:"Maybe", Nothing:false, Just:x}
end Just

-- Nothing :: Maybe a
on Nothing()
    {type:"Maybe", Nothing:true}
end Nothing

-- enumFromTo :: (Int, Int) -> [Int]
on enumFromTo(m, n)
    if m ≤ n then
        set lst to {}
        repeat with i from m to n
            set end of lst to i
        end repeat
        return lst
    else
        return {}
    end if
end enumFromTo

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl

-- index (!!) :: [a] -> Int -> Maybe a
-- index (!!) :: Gen [a] -> Int -> Maybe a
-- index (!!) :: String -> Int -> Maybe Char
on |index|(xs, i)
    if script is class of xs then
        repeat with j from 1 to i
            set v to |λ|() of xs
        end repeat
        if missing value is not v then
            Just(v)
        else
            Nothing()
        end if
    else
        if length of xs < i then
            Nothing()
        else
            Just(item i of xs)
        end if
    end if
end |index|

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map

-- min :: Ord a => a -> a -> a
on min(x, y)
    if y < x then
        y
    else
        x
    end if
end min

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn

-- repeat :: a -> Generator [a]
on |repeat|(x)
    script
        on |λ|()
            return x
        end |λ|
    end script
end |repeat|


-- scanl :: (b -> a -> b) -> b -> [a] -> [b]
on scanl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        set lst to {startValue}
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
            set end of lst to v
        end repeat
        return lst
    end tell
end scanl

-- scanl1 :: (a -> a -> a) -> [a] -> [a]
on scanl1(f, xs)
    if 0 < length of xs then
        scanl(f, item 1 of xs, rest of xs)
    else
        {}
    end if
end scanl1


-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
    set c to class of xs
    if list is c then
        if 0 < n then
            items 1 thru min(n, length of xs) of xs
        else
            {}
        end if
    else if string is c then
        if 0 < n then
            text 1 thru min(n, length of xs) of xs
        else
            ""
        end if
    else if script is c then
        set ys to {}
        repeat with i from 1 to n
            set v to |λ|() of xs
            if missing value is v then
                return ys
            else
                set end of ys to v
            end if
        end repeat
        return ys
    else
        missing value
    end if
end take

{220, {{"iced", "iced"}, {"jam", "iced"}, {"jam", "jam"}, {"plain", "iced"}, {"plain", "jam"}, {"plain", "plain"}}}

AWK


# syntax: GAWK -f COMBINATIONS_WITH_REPETITIONS.AWK
BEGIN {
    n = split("iced,jam,plain",donuts,",")
    for (i=1; i<=n; i++) {
      for (j=1; j<=n; j++) {
        if (donuts[i] < donuts[j]) {
          key = sprintf("%s %s",donuts[i],donuts[j])
        }
        else {
          key = sprintf("%s %s",donuts[j],donuts[i])
        }
        arr[key]++
      }
    }
    cmd = "SORT"
    for (i in arr) {
      printf("%s\n",i) | cmd
      choices++
    }
    close(cmd)
    printf("choices = %d\n",choices)
    exit(0)
}

output:


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
choices = 6

BBC BASIC

      DIM list$(2), chosen%(2)
      list$() = "iced", "jam", "plain"
      PRINT "Choices of 2 from 3:"
      choices% = FNchoose(0, 2, 0, 3, chosen%(), list$())
      PRINT "Total choices = " ; choices%

      PRINT '"Choices of 3 from 10:"
      choices% = FNchoose(0, 3, 0, 10, chosen%(), nul$())
      PRINT "Total choices = " ; choices%
      END

      DEF FNchoose(n%, l%, p%, m%, g%(), RETURN n$())
      LOCAL i%, c%
      IF n% = l% THEN
        IF !^n$() THEN
          FOR i% = 0 TO n%-1
            PRINT " " n$(g%(i%)) ;
          NEXT
          PRINT
        ENDIF
        = 1
      ENDIF
      FOR i% = p% TO m%-1
        g%(n%) = i%
        c% += FNchoose(n% + 1, l%, i%, m%, g%(), n$())
      NEXT
      = c%


Choices of 2 from 3:
 iced iced
 iced jam
 iced plain
 jam jam
 jam plain
 plain plain
Total choices = 6

Choices of 3 from 10:
Total choices = 220

Bracmat

This minimalist solution expresses the answer as a sum of products. Bracmat automatically normalises such expressions: terms and factors are sorted alphabetically, products containing a sum as a factor are decomposed in a sum of factors (unless the product is not itself term in a multiterm expression). Like factors are converted to a single factor with an appropriate exponent, so ice^2 is to be understood as ice twice.

( ( choices
  =   n things thing result
    .   !arg:(?n.?things)
      & ( !n:0&1
        |   0:?result
          & (   !things
              :   ?
                  ( %?`thing ?:?things
                  &   !thing*choices$(!n+-1.!things)+!result
                    : ?result
                  & ~
                  )
            | !result
            )
        )
  )
& out$(choices$(2.iced jam plain))
& out$(choices$(3.iced jam plain butter marmite tahin fish salad onion grass):?+[?N&!N)
);

iced^2+jam^2+plain^2+iced*jam+iced*plain+jam*plain
220

C

#include <stdio.h>

const char * donuts[] = { "iced", "jam", "plain", "something completely different" };
long choose(int * got, int n_chosen, int len, int at, int max_types)
{
        int i;
        long count = 0;
        if (n_chosen == len) {
                if (!got) return 1;

                for (i = 0; i < len; i++)
                        printf("%s\t", donuts[got[i]]);
                printf("\n");
                return 1;
        }

        for (i = at; i < max_types; i++) {
                if (got) got[n_chosen] = i;
                count += choose(got, n_chosen + 1, len, i, max_types);
        }
        return count;
}

int main()
{
        int chosen[3];
        choose(chosen, 0, 2, 0, 3);

        printf("\nWere there ten donuts, we'd have had %ld choices of three\n",
                choose(0, 0, 3, 0, 10));
        return 0;
}

iced    iced
iced    jam
iced    plain
jam     jam
jam     plain
plain   plain

Were there ten donuts, we'd have had 220 choices of three

C++

Non recursive version.


#include <cstdio>
#include <vector>
#include <string>

using namespace std;

void print_vector(const vector<int> &v, size_t n, const vector<string> &s){
        for (size_t i = 0; i < n; ++i)
                printf("%s\t", s[v[i]].c_str());
        printf("\n");
}

void combination_with_repetiton(int sabores, int bolas, const vector<string>& v_sabores){
        sabores--;
        vector<int> v(bolas+1, 0);
        while (true){
                for (int i = 0; i < bolas; ++i){                //vai um
                        if (v[i] > sabores){
                                v[i + 1] += 1;
                                for (int k = i; k >= 0; --k){
                                        v[k] = v[i + 1];
                                }
                                //v[i] = v[i + 1];
                        }
                }

                if (v[bolas] > 0)
                        break;
                print_vector(v, bolas, v_sabores);
                v[0] += 1;
        }
}

int main(){
        vector<string> options{ "iced", "jam", "plain" };
        combination_with_repetiton(3, 2, options);
        return 0;
}


iced	iced
jam	iced
plain	iced
jam	jam
plain	jam
plain	plain

C#


using System;
using System.Collections.Generic;
using System.Linq;

public static class MultiCombinations
{
    private static void Main()
    {
        var set = new List<string> { "iced", "jam", "plain" };
        var combinations = GenerateCombinations(set, 2);

        foreach (var combination in combinations)
        {
            string combinationStr = string.Join(" ", combination);
            Console.WriteLine(combinationStr);
        }

        var donuts = Enumerable.Range(1, 10).ToList();

        int donutsCombinationsNumber = GenerateCombinations(donuts, 3).Count;

        Console.WriteLine("{0} ways to order 3 donuts given 10 types", donutsCombinationsNumber);
    }

    private static List<List<T>> GenerateCombinations<T>(List<T> combinationList, int k)
    {
        var combinations = new List<List<T>>();

        if (k == 0)
        {
            var emptyCombination = new List<T>();
            combinations.Add(emptyCombination);

            return combinations;
        }

        if (combinationList.Count == 0)
        {
            return combinations;
        }

        T head = combinationList[0];
        var copiedCombinationList = new List<T>(combinationList);

        List<List<T>> subcombinations = GenerateCombinations(copiedCombinationList, k - 1);

        foreach (var subcombination in subcombinations)
        {
            subcombination.Insert(0, head);
            combinations.Add(subcombination);
        }

        combinationList.RemoveAt(0);
        combinations.AddRange(GenerateCombinations(combinationList, k));

        return combinations;
    }
}


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
220 ways to order 3 donuts given 10 types

Recursive version


using System;
class MultiCombination
{
  static string [] set = { "iced", "jam", "plain" };
  static int k = 2, n = set.Length;
  static string [] buf = new string [k];

  static void Main()
  {
    rec(0, 0);
  }

  static void rec(int ind, int begin)
  {
    for (int i = begin; i < n; i++)
    {
      buf [ind] = set[i];
      if (ind + 1 < k) rec(ind + 1, i);
      else Console.WriteLine(string.Join(",", buf));
    }
  }
}

Clojure


(defn combinations [coll k]
  (when-let [[x & xs] coll]
    (if (= k 1)
      (map list coll)
      (concat (map (partial cons x) (combinations coll (dec k)))
              (combinations xs k)))))


user> (combinations '[iced jam plain] 2)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

CoffeeScript


combos = (arr, k) ->
  return [ [] ] if k == 0
  return [] if arr.length == 0

  combos_with_head = ([arr[0]].concat combo for combo in combos arr, k-1)
  combos_sans_head = combos arr[1...], k
  combos_with_head.concat combos_sans_head

arr = ['iced', 'jam', 'plain']
console.log "valid pairs from #{arr.join ','}:"
console.log combos arr, 2
console.log "#{combos([1..10], 3).length} ways to order 3 donuts given 10 types"


jam,plain:
[ [ 'iced', 'iced' ],
  [ 'iced', 'jam' ],
  [ 'iced', 'plain' ],
  [ 'jam', 'jam' ],
  [ 'jam', 'plain' ],
  [ 'plain', 'plain' ] ]
220 ways to order 3 donuts given 10 types

Common Lisp

The code below is a modified version of the Clojure solution.

(defun combinations (xs k)
  (let ((x (car xs)))
    (cond
     ((null xs) nil)
     ((= k 1) (mapcar #'list xs))
     (t (append (mapcar (lambda (ys) (cons x ys))
			(combinations xs (1- k)))
		(combinations (cdr xs) k))))))

((:ICED :ICED) (:ICED :JAM) (:ICED :PLAIN) (:JAM :JAM) (:JAM :PLAIN) (:PLAIN :PLAIN))

Crystal

possible_doughnuts = ["iced", "jam", "plain"].repeated_combinations(2)
puts "There are #{possible_doughnuts.size} possible doughnuts:"
possible_doughnuts.each{|doughnut_combi| puts doughnut_combi.join(" and ")}

# Extra credit
possible_doughnuts = (1..10).to_a.repeated_combinations(3)
# size returns the size of the enumerator, or nil if it can’t be calculated lazily.
puts "", "#{possible_doughnuts.size} ways to order 3 donuts given 10 types."


There are 6 possible doughnuts:
iced and iced
iced and jam
iced and plain
jam and jam
jam and plain
plain and plain

220 ways to order 3 donuts given 10 types.

D

Using [http://www.graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation lexicographic next bit permutation] to generate combinations with repetitions.

import std.stdio, std.range;

const struct CombRep {
    immutable uint nt, nc;
    private const ulong[] combVal;

    this(in uint numType, in uint numChoice) pure nothrow @safe
    in {
        assert(0 < numType && numType + numChoice <= 64,
               "Valid only for nt + nc <= 64 (ulong bit size)");
    } body {
        nt = numType;
        nc = numChoice;
        if (nc == 0)
            return;
        ulong v  = (1UL << (nt - 1)) - 1;

        // Init to smallest number that has nt-1 bit set
        // a set bit is metaphored as a _type_ seperator.
        immutable limit = v << nc;

        ulong[] localCombVal;
        // Limit is the largest nt-1 bit set number that has nc
        // zero-bit a zero-bit means a _choice_ between _type_
        // seperators.
        while (v <= limit) {
            localCombVal ~= v;
            if (v == 0)
                break;
            // Get next nt-1 bit number.
            immutable t = (v | (v - 1)) + 1;
            v = t | ((((t & -t) / (v & -v)) >> 1) - 1);
        }
        this.combVal = localCombVal;
    }

    uint length() @property const pure nothrow @safe {
        return combVal.length;
    }

    uint[] opIndex(in uint idx) const pure nothrow @safe {
        return val2set(combVal[idx]);
    }

    int opApply(immutable int delegate(in ref uint[]) pure nothrow @safe dg)
    pure nothrow @safe {
        foreach (immutable v; combVal) {
            auto set = val2set(v);
            if (dg(set))
                break;
        }
        return 1;
    }

    private uint[] val2set(in ulong v) const pure nothrow @safe {
        // Convert bit pattern to selection set
        immutable uint bitLimit = nt + nc - 1;
        uint typeIdx = 0;
        uint[] set;
        foreach (immutable bitNum; 0 .. bitLimit)
            if (v & (1 << (bitLimit - bitNum - 1)))
                typeIdx++;
            else
                set ~= typeIdx;
        return set;
    }
}

// For finite Random Access Range.
auto combRep(R)(R types, in uint numChoice) /*pure*/ nothrow @safe
if (hasLength!R && isRandomAccessRange!R) {
    ElementType!R[][] result;

    foreach (const s; CombRep(types.length, numChoice)) {
        ElementType!R[] r;
        foreach (immutable i; s)
            r ~= types[i];
        result ~= r;
    }

    return result;
}

void main() @safe {
    foreach (const e; combRep(["iced", "jam", "plain"], 2))
        writefln("%-(%5s %)", e);
    writeln("Ways to select 3 from 10 types is ",
            CombRep(10, 3).length);
}

 iced  iced
 iced   jam
 iced plain
  jam   jam
  jam plain
plain plain
Ways to select 3 from 10 types is 220

Short Recursive Version

import std.stdio, std.range, std.algorithm;

T[][] combsRep(T)(T[] lst, in int k) {
    if (k == 0)
        return [[]];
    if (lst.empty)
        return null;

    return combsRep(lst, k - 1).map!(L => lst[0] ~ L).array
           ~ combsRep(lst[1 .. $], k);
}

void main() {
    ["iced", "jam", "plain"].combsRep(2).writeln;
    10.iota.array.combsRep(3).length.writeln;
}

[["iced", "iced"], ["iced", "jam"], ["iced", "plain"], ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]]
220

EasyLang

items$[] = [ "iced" "jam" "plain" ] n = len items$[] k = 2 len result[] k n_results = 0

func output . . n_results += 1 if len items$[] > 0 s$ = "" for i range k s$ &= items$[result[i]] & " " . print s$ . . func combine pos val . . if pos = k call output else for i = val to n - 1 result[pos] = i call combine pos + 1 i . . . call combine 0 0

n = 10 k = 3 len result[] k items$[] = [ ] n_results = 0 call combine 0 0 print "" print n_results & " results with 10 donuts"



{{out}}

```txt

iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

220 results with 10 donuts

EchoLisp

We can use the native '''combinations/rep''' function, or use a '''combinator''' iterator, or implement the function.


;;
;; native function : combinations/rep in list.lib
;;
(lib 'list)

(combinations/rep '(iced jam plain) 2)
   → ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; using a combinator iterator
;;
(lib 'sequences)

(take (combinator/rep '(iced jam plain) 2) 8)
    → ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; or, implementing the function
;;

(define (comb/rep nums k)
	(cond
	[(null? nums) null]
	[(<= k 0) null]
	[(= k 1) (map list nums)]
	[else
		(for/fold (acc null) ((anum nums))
		(append acc
	  		(for/list ((xs (comb/rep nums (1- k))))
	  		#:continue (< (first xs) anum)
	  		(cons anum xs))))]))

(map (curry list-permute '(iced jam plain)) (comb/rep (iota 3) 2))
    → ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; extra credit
;;

(length (combinator/rep (iota 10) 3))
    → 220

Egison


(define $comb/rep
  (lambda [$n $xs]
    (match-all xs (list something)
      [(loop $i [1 ,n] <join _ (& <cons $a_i _> ...)> _) a])))

(test (comb/rep 2 {"iced" "jam" "plain"}))


{[|"iced" "iced"|] [|"iced" "jam"|] [|"jam" "jam"|] [|"iced" "plain"|] [|"jam" "plain"|] [|"plain" "plain"|]}

Elixir

defmodule RC do
  def comb_rep(0, _), do: [[]]
  def comb_rep(_, []), do: []
  def comb_rep(n, [h|t]=s) do
    (for l <- comb_rep(n-1, s), do: [h|l]) ++ comb_rep(n, t)
  end
end

s = [:iced, :jam, :plain]
Enum.each(RC.comb_rep(2, s), fn x -> IO.inspect x end)

IO.puts  "\nExtra credit: #{length(RC.comb_rep(3, Enum.to_list(1..10)))}"


[:iced, :iced]
[:iced, :jam]
[:iced, :plain]
[:jam, :jam]
[:jam, :plain]
[:plain, :plain]

Extra credit: 220

Erlang


-module(comb).
-compile(export_all).

comb_rep(0,_) ->
    [[]];
comb_rep(_,[]) ->
    [];
comb_rep(N,[H|T]=S) ->
    [[H|L] || L <- comb_rep(N-1,S)]++comb_rep(N,T).


94> comb:comb_rep(2,[iced,jam,plain]).
[[iced,iced],
 [iced,jam],
 [iced,plain],
 [jam,jam],
 [jam,plain],
 [plain,plain]]
95> length(comb:comb_rep(3,lists:seq(1,10))).
220

Fortran


program main
	integer :: chosen(4)
	integer :: ssize

	character(len=50) :: donuts(4) = [ "iced", "jam", "plain", "something completely different" ]

	ssize = choose( chosen, 2, 3 )
	write(*,*) "Total = ", ssize

	contains

	recursive function choose( got, len, maxTypes, nChosen, at ) result ( output )
		integer :: got(:)
		integer :: len
		integer :: maxTypes
		integer :: output
		integer, optional :: nChosen
		integer, optional :: at

		integer :: effNChosen
		integer :: effAt

		integer :: i
		integer :: counter

		effNChosen = 1
		if( present(nChosen) ) effNChosen = nChosen

		effAt = 1
		if( present(at) ) effAt = at

		if ( effNChosen == len+1 ) then
			do i=1,len
				write(*,"(A10,5X)", advance='no') donuts( got(i)+1 )
			end do

			write(*,*) ""

			output = 1
			return
		end if

		counter = 0
		do i=effAt,maxTypes
			got(effNChosen) = i-1
			counter = counter + choose( got, len, maxTypes, effNChosen + 1, i )
		end do

		output = counter
		return
	end function choose

end program main


iced           iced
iced           jam
iced           plain
jam            jam
jam            plain
plain          plain
 Total =            6

GAP

# Built-in
UnorderedTuples(["iced", "jam", "plain"], 2);

Go

Concise recursive

package main

import "fmt"

func combrep(n int, lst []string) [][]string {
    if n == 0 {
        return [][]string{nil}
    }
    if len(lst) == 0 {
        return nil
    }
    r := combrep(n, lst[1:])
    for _, x := range combrep(n-1, lst) {
        r = append(r, append(x, lst[0]))
    }
    return r
}

func main() {
    fmt.Println(combrep(2, []string{"iced", "jam", "plain"}))
    fmt.Println(len(combrep(3,
        []string{"1", "2", "3", "4", "5", "6", "7", "8", "9", "10"})))
}


[[plain plain] [plain jam] [jam jam] [plain iced] [jam iced] [iced iced]]
220

Channel

Using channel and goroutine, showing how to use synced or unsynced communication.

package main

import "fmt"

func picks(picked []int, pos, need int, c chan[]int, do_wait bool) {
	if need == 0 {
		if do_wait {
			c <- picked
			<-c
		} else { // if we want only the count, there's no need to
			 // sync between coroutines; let it clobber the array
			c <- []int {}
		}
		return
	}

	if pos <= 0 {
		if need == len(picked) { c <- nil }
		return
	}

	picked[len(picked) - need] = pos - 1
	picks(picked, pos, need - 1, c, do_wait) // choose the current donut
	picks(picked, pos - 1, need, c, do_wait) // or don't
}

func main() {
	donuts := []string {"iced", "jam", "plain" }

	picked := make([]int, 2)
	ch := make(chan []int)

	// true: tell the channel to wait for each sending, because
	// otherwise the picked array may get clobbered before we can do
	// anything to it
	go picks(picked, len(donuts), len(picked), ch, true)

	var cc []int
	for {
		if cc = <-ch; cc == nil { break }
		for _, i := range cc {
			fmt.Printf("%s ", donuts[i])
		}
		fmt.Println()
		ch <- nil // sync
	}

	picked = make([]int, 3)
	// this time we only want the count, so tell goroutine to keep going
	// and work the channel buffer
	go picks(picked, 10, len(picked), ch, false)
	count := 0
	for {
		if cc = <-ch; cc == nil { break }
		count++
	}
	fmt.Printf("\npicking 3 of 10: %d\n", count)
}


plain plain
plain jam
plain iced
jam jam
jam iced
iced iced

picking 3 of 10: 220

Multiset

This version has proper representation of sets and multisets.

package main

import (
    "fmt"
    "sort"
    "strconv"
)

// Go maps are an easy representation for sets as long as the element type
// of the set is valid as a key type for maps.  Strings are easy.
// We follow the convention of always storing true for the value.
type set      map[string]bool

// Multisets of strings are easy in the same way.
// We store the multiplicity of the element as the value.
type multiset map[string]int

// But multisets are not valid as a map key type so we must do something
// more involved to make a set of multisets, which is the desired return
// type for the combrep function required by the task.  We can store the
// multiset as the value, but we derive a unique string to use as a key.
type msSet    map[string]multiset

// The key method returns this string.  The string will simply be a text
// representation of the contents of the multiset.  The standard
// printable representation of the multiset cannot be used however, because
// Go maps are not ordered.  Instead, the contents are copied to a slice,
// which is sorted to produce something with a printable representation
// that will compare == for mathematically equal multisets.
//
// Of course there is overhead for this and if performance were important,
// a different representation would be used for multisets, one that didn’t
// require sorting to produce a key...
func (m multiset) key() string {
    pl := make(pairList, len(m))
    i := 0
    for k, v := range m {
        pl[i] = msPair{k, v}
	i++
    }
    sort.Sort(pl)
    return fmt.Sprintf("%v", pl)
}

// Types and methods needed for sorting inside of mulitset.key()
type msPair struct {
    string
    int
}
type pairList []msPair
func (p pairList) Len() int { return len(p) }
func (p pairList) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
func (p pairList) Less(i, j int) bool { return p[i].string < p[j].string }

// Function required by task.
func combrep(n int, lst set) msSet {
    if n == 0 {
        var ms multiset
        return msSet{ms.key(): ms}
    }
    if len(lst) == 0 {
        return msSet{}
    }
    var car string
    var cdr set
    for ele := range lst {
        if cdr == nil {
            car = ele
            cdr = make(set)
        } else {
            cdr[ele] = true
        }
    }
    r := combrep(n, cdr)

    for _, x := range combrep(n-1, lst) {
        c := multiset{car: 1}
        for k, v := range x {
            c[k] += v
        }
        r[c.key()] = c
    }
    return r
}

// Driver for examples required by task.
func main() {
    // Input is a set.
    three := set{"iced": true, "jam": true, "plain": true}
    // Output is a set of multisets.  The set is a Go map:
    // The key is a string representation that compares equal
    // for equal multisets.  We ignore this here.  The value
    // is the multiset.  We print this.
    for _, ms := range combrep(2, three) {
        fmt.Println(ms)
    }
    ten := make(set)
    for i := 1; i <= 10; i++ {
        ten[strconv.Itoa(i)] = true
    }
    fmt.Println(len(combrep(3, ten)))
}


map[plain:1 jam:1]
map[plain:2]
map[iced:1 jam:1]
map[jam:2]
map[iced:1 plain:1]
map[iced:2]
220

Haskell

-- Return the combinations, with replacement, of k items from the
-- list.  We ignore the case where k is greater than the length of
-- the list.
combsWithRep :: Int -> [a] -> [[a]]
combsWithRep 0 _ = [[]]
combsWithRep _ [] = []
combsWithRep k xxs@(x:xs) =
  (x :) <$> combsWithRep (k - 1) xxs ++ combsWithRep k xs

binomial n m = f n `div` f (n - m) `div` f m
  where
    f n =
      if n == 0
        then 1
        else n * f (n - 1)

countCombsWithRep :: Int -> [a] -> Int
countCombsWithRep k lst = binomial (k - 1 + length lst) k

-- countCombsWithRep k = length . combsWithRep k
main :: IO ()
main = do
  print $ combsWithRep 2 ["iced", "jam", "plain"]
  print $ countCombsWithRep 3 [1 .. 10]

[["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"],["jam","plain"],["plain","plain"]]
220

Dynamic Programming

The first solution is inefficient because it repeatedly calculates the same subproblem in different branches of recursion. For example, combsWithRep k (x:xs) involves computing combsWithRep (k-1) (x:xs) and combsWithRep k xs, both of which (separately) compute combsWithRep (k-1) xs. To avoid repeated computation, we can use dynamic programming:

combsWithRep :: Int -> [a] -> [[a]]
combsWithRep k xs = combsBySize xs !! k
  where
    combsBySize = foldr f ([[]] : repeat [])
    f x = scanl1 $ (++) . map (x :)

main :: IO ()
main = print $ combsWithRep 2 ["iced", "jam", "plain"]

and another approach, using manual recursion:

combsWithRep
  :: (Eq a)
  => Int -> [a] -> [[a]]
combsWithRep k xs = comb k []
  where
    comb 0 lst = lst
    comb n [] = comb (n - 1) (pure <$> xs)
    comb n peers =
      let nextLayer ys@(h:_) = (: ys) <$> dropWhile (/= h) xs
      in comb (n - 1) (foldMap nextLayer peers)

main :: IO ()
main = do
  print $ combsWithRep 2 ["iced", "jam", "plain"]
  print $ length $ combsWithRep 3 [0 .. 9]

[["iced","iced"],["jam","iced"],["plain","iced"],["jam","jam"],["plain","jam"],["plain","plain"]]
220

=={{header|Icon}} and {{header|Unicon}}==

Following procedure is a generator, which generates each combination of length n in turn:


# generate all combinations of length n from list L,
# including repetitions
procedure combinations_repetitions (L, n)
  if n = 0
    then suspend [] # if reach 0, then return an empty list
    else if *L > 0
      then {
        # keep the first element
        item := L[1]
        # get all of length n in remaining list
        every suspend (combinations_repetitions (L[2:0], n))
        # get all of length n-1 in remaining list
        # and add kept element to make list of size n
        every i := combinations_repetitions (L, n-1) do
          suspend [item] ||| i
      }
end

Test procedure:


# convenience function
procedure write_list (l)
  every (writes (!l || " "))
  write ()
end

# testing routine
procedure main ()
  # display all combinations for 2 of iced/jam/plain
  every write_list (combinations_repetitions(["iced", "jam", "plain"], 2))
  # get a count for number of ways to select 3 items from 10
  every push(num_list := [], 1 to 10)
  count := 0
  every combinations_repetitions(num_list, 3) do count +:= 1
  write ("There are " || count || " possible combinations of 3 from 10")
end


plain plain
jam plain
jam jam
iced plain
iced jam
iced iced
There are 220 possible combinations of 3 from 10

=={{header|IS-BASIC}}== 100 PROGRAM "Combinat.bas" 110 READ N 120 STRING D$(1 TO N)*5 130 FOR I=1 TO N 140 READ D$(I) 150 NEXT 160 FOR I=1 TO N 170 FOR J=I TO N 180 PRINT D$(I);" ";D$(J) 190 NEXT 200 NEXT 210 DATA 3,iced,jam,plain




## J

Cartesian product, the monadic j verb { solves the problem.  The rest of the code handles the various data types, order, and quantity to choose, and makes a set from the result.


```j>rcomb=:
@~.@:(/:~&.>)@,@{@# <

Example use:

   2 rcomb ;:'iced jam plain'
┌─────┬─────┐
│iced │iced │
├─────┼─────┤
│iced │jam  │
├─────┼─────┤
│iced │plain│
├─────┼─────┤
│jam  │jam  │
├─────┼─────┤
│jam  │plain│
├─────┼─────┤
│plain│plain│
└─────┴─────┘
   #3 rcomb i.10         NB. # ways to choose 3 items from 10 with repetitions
220

J Alternate implementation

Considerably faster:

require 'stats'
combr=: i.@[ -"1~ [ comb + - 1:
rcomb=: (combr #) { ]

rcomb functions identically, and combr calculates indices:

In other words: we compute 2 comb 4 (note that 4 = (2 + 3)-1) and then subtract from each column the minimum value in each column (i. 2).

Java

'''MultiCombinationsTester.java'''


import com.objectwave.utility.*;

public class MultiCombinationsTester {

    public MultiCombinationsTester() throws CombinatoricException {
        Object[] objects = {"iced", "jam", "plain"};
        //Object[] objects = {"abba", "baba", "ab"};
        //Object[] objects = {"aaa", "aa", "a"};
        //Object[] objects = {(Integer)1, (Integer)2, (Integer)3, (Integer)4};
        MultiCombinations mc = new MultiCombinations(objects, 2);
        while (mc.hasMoreElements()) {
            for (int i = 0; i < mc.nextElement().length; i++) {
                System.out.print(mc.nextElement()[i].toString() + " ");
            }
            System.out.println();
        }

        // Extra credit:
        System.out.println("----------");
        System.out.println("The ways to choose 3 items from 10 with replacement = " + MultiCombinations.c(10, 3));
    } // constructor

    public static void main(String[] args) throws CombinatoricException {
        new MultiCombinationsTester();
    }
} // class

'''MultiCombinations.java'''


import com.objectwave.utility.*;
import java.util.*;

public class MultiCombinations {

    private HashSet<String> set = new HashSet<String>();
    private Combinations comb = null;
    private Object[] nextElem = null;

    public MultiCombinations(Object[] objects, int k) throws CombinatoricException {
        k = Math.max(0, k);
        Object[] myObjects = new Object[objects.length * k];
        for (int i = 0; i < objects.length; i++) {
            for (int j = 0; j < k; j++) {
                myObjects[i * k + j] = objects[i];
            }
        }
        comb = new Combinations(myObjects, k);
    } // constructor

    boolean hasMoreElements() {
        boolean ret = false;
        nextElem = null;
        int oldCount = set.size();
        while (comb.hasMoreElements()) {
            Object[] elem = (Object[]) comb.nextElement();
            String str = "";
            for (int i = 0; i < elem.length; i++) {
                str += ("%" + elem[i].toString() + "~");
            }
            set.add(str);
            if (set.size() > oldCount) {
                nextElem = elem;
                ret = true;
                break;
            }
        }
        return ret;
    } // hasMoreElements()

    Object[] nextElement() {
        return nextElem;
    }

    static java.math.BigInteger c(int n, int k) throws CombinatoricException {
        return Combinatoric.c(n + k - 1, k);
    }
} // class


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
----------
The ways to choose 3 items from 10 with replacement = 220

JavaScript

ES5

=Imperative=


<body><pre id='x'>


{{out}}

```txt
iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
6 combos
pick 3 out of 10: 220 combos

=Functional=

(function () {

  // n -> [a] -> [[a]]
  function combsWithRep(n, lst) {
    return n ? (
      lst.length ? combsWithRep(n - 1, lst).map(function (t) {
        return [lst[0]].concat(t);
      }).concat(combsWithRep(n, lst.slice(1))) : []
    ) : [[]];
  };

  // If needed, we can derive a significantly faster version of
  // the simple recursive function above by memoizing it

  // f -> f
  function memoized(fn) {
    m = {};
    return function (x) {
      var args = [].slice.call(arguments),
        strKey = args.join('-');

      v = m[strKey];
      if ('u' === (typeof v)[0])
        m[strKey] = v = fn.apply(null, args);
      return v;
    }
  }

  // [m..n]
  function range(m, n) {
    return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
      return m + i;
    });
  }


  return [

      combsWithRep(2, ["iced", "jam", "plain"]),

    // obtaining and applying a memoized version of the function
      memoized(combsWithRep)(3, range(1, 10)).length
    ];

})();

[
 [["iced", "iced"], ["iced", "jam"], ["iced", "plain"],
  ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]],
 220
]

ES6

(() => {
    'use strict';

    // COMBINATIONS WITH REPETITIONS -------------------------------------------

    // combsWithRep :: Int -> [a] -> [[a]]
    const combsWithRep = (k, xs) => {
        const comb = (n, ys) => {
            if (0 === n) return ys;
            if (isNull(ys)) return comb(n - 1, map(pure, xs));

            return comb(n - 1, concatMap(zs => {
                const h = head(zs);
                return map(x => [x].concat(zs), dropWhile(x => x !== h, xs));
            }, ys));
        };
        return comb(k, []);
    };

    // GENERIC FUNCTIONS ------------------------------------------------------

    // concatMap :: (a -> [b]) -> [a] -> [b]
    const concatMap = (f, xs) => [].concat.apply([], xs.map(f));

    // dropWhile :: (a -> Bool) -> [a] -> [a]
    const dropWhile = (p, xs) => {
        let i = 0;
        for (let lng = xs.length;
            (i < lng) && p(xs[i]); i++) {}
        return xs.slice(i);
    };

    // enumFromTo :: Int -> Int -> [Int]
    const enumFromTo = (m, n) =>
        Array.from({
            length: Math.floor(n - m) + 1
        }, (_, i) => m + i);

    // head :: [a] -> Maybe a
    const head = xs => xs.length ? xs[0] : undefined;

    // isNull :: [a] -> Bool
    const isNull = xs => (xs instanceof Array) ? xs.length < 1 : undefined;

    // length :: [a] -> Int
    const length = xs => xs.length;

    // map :: (a -> b) -> [a] -> [b]
    const map = (f, xs) => xs.map(f);

    // pure :: a -> [a]
    const pure = x => [x];

    // show :: a -> String
    const show = x => JSON.stringify(x, null, 2);

    // TEST -------------------------------------------------------------------
    return show({
        twoFromThree: combsWithRep(2, ['iced', 'jam', 'plain']),
        threeFromTen: length(combsWithRep(3, enumFromTo(0, 9)))
    });
})();

{
  "twoFromThree": [
    [
      "iced",
      "iced"
    ],
    [
      "jam",
      "iced"
    ],
    [
      "plain",
      "iced"
    ],
    [
      "jam",
      "jam"
    ],
    [
      "plain",
      "jam"
    ],
    [
      "plain",
      "plain"
    ]
  ],
  "threeFromTen": 220
}

jq

def pick(n):
  def pick(n; m):  # pick n, from m onwards
    if n == 0 then []
    elif m == length then empty
    elif n == 1 then (.[m:][] | [.])
    else ([.[m]] + pick(n-1; m)), pick(n; m+1)
    end;
  pick(n;0) ;

'''The task''':

 "Pick 2:",
 (["iced", "jam", "plain"] | pick(2)),
 ([[range(0;10)] | pick(3)] | length) as $n
  | "There are \($n) ways to pick 3 objects with replacement from 10."

$ jq -n -r -c -f pick.jq
Pick 2:
["iced","iced"]
["iced","jam"]
["iced","plain"]
["jam","jam"]
["jam","plain"]
["plain","plain"]
There are 220 ways to pick 3 objects with replacement from 10.

Julia

using Combinatorics

l = ["iced", "jam", "plain"]
println("List: ", l, "\nCombinations:")
for c in with_replacement_combinations(l, 2)
    println(c)
end

@show length(with_replacement_combinations(1:10, 3))

List: String["iced", "jam", "plain"]
Combinations:
String["iced", "iced"]
String["iced", "jam"]
String["iced", "plain"]
String["jam", "jam"]
String["jam", "plain"]
String["plain", "plain"]
length(with_replacement_combinations(1:10, 3)) = 220

Kotlin

// version 1.0.6

class CombsWithReps<T>(val m: Int, val n: Int, val items: List<T>, val countOnly: Boolean = false) {
    private val combination = IntArray(m)
    private var count = 0

    init {
        generate(0)
        if (!countOnly) println()
        println("There are $count combinations of $n things taken $m at a time, with repetitions")
    }

    private fun generate(k: Int) {
        if (k >= m) {
            if (!countOnly) {
                for (i in 0 until m) print("${items[combination[i]]}\t")
                println()
            }
            count++
        }
        else {
            for (j in 0 until n)
                if (k == 0 || j >= combination[k - 1]) {
                    combination[k] = j
                    generate(k + 1)
                }
        }
    }
}

fun main(args: Array<String>) {
    val doughnuts = listOf("iced", "jam", "plain")
    CombsWithReps(2, 3, doughnuts)
    println()
    val generic10 = "0123456789".split("")
    CombsWithReps(3, 10, generic10, true)
}


iced    iced
iced    jam
iced    plain
jam     jam
jam     plain
plain   plain

There are 6 combinations of 3 things taken 2 at a time, with repetitions

There are 220 combinations of 10 things taken 3 at a time, with repetitions

LFE

{{trans|Erlang}} and {{trans|Clojure}}

With List Comprehension


(defun combinations
 (('() _)
    '())
  ((coll 1)
    (lists:map #'list/1 coll))
   (((= (cons head tail) coll) n)
    (++ (lc ((<- subcoll (combinations coll (- n 1))))
            (cons head subcoll))
        (combinations tail n))))

With Map


(defun combinations
  (('() _)
    '())
  ((coll 1)
    (lists:map #'list/1 coll))
   (((= (cons head tail) coll) n)
    (++ (lists:map (lambda (subcoll) (cons head subcoll))
                   (combinations coll (- n 1)))
        (combinations tail n))))

Output is the same for both:


> (combinations '(iced jam plain) 2)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

Lua

function GenerateCombinations(tList, nMaxElements, tOutput, nStartIndex, nChosen, tCurrentCombination)
	if not nStartIndex then
		nStartIndex = 1
	end
	if not nChosen then
		nChosen = 0
	end
	if not tOutput then
		tOutput = {}
	end
	if not tCurrentCombination then
		tCurrentCombination = {}
	end

	if nChosen == nMaxElements then
		-- Must copy the table to avoid all elements referring to a single reference
		local tCombination = {}
		for k,v in pairs(tCurrentCombination) do
			tCombination[k] = v
		end

		table.insert(tOutput, tCombination)
		return
	end

 	local nIndex = 1
	for k,v in pairs(tList) do
		if nIndex >= nStartIndex then
			tCurrentCombination[nChosen + 1] = tList[nIndex]
			GenerateCombinations(tList, nMaxElements, tOutput, nIndex, nChosen + 1, tCurrentCombination)
		end

		nIndex = nIndex + 1
	end

	return tOutput
end

tDonuts = {"iced", "jam", "plain"}
tCombinations = GenerateCombinations(tDonuts, #tDonuts)
for nCombination,tCombination in ipairs(tCombinations) do
	print("Combination " .. tostring(nCombination))
	for nIndex,strFlavor in ipairs(tCombination) do
		print("+" .. strFlavor)
	end
end

=={{header|Mathematica}} / {{header|Wolfram Language}}==

This method will only work for small set and sample sizes (as it generates all Tuples then filters duplicates - Length[Tuples[Range[10],10]] is already bigger than Mathematica can handle).

DeleteDuplicates[Tuples[{"iced", "jam", "plain"}, 2],Sort[#1] == Sort[#2] &]
->{{"iced", "iced"}, {"iced", "jam"}, {"iced", "plain"}, {"jam", "jam"}, {"jam", "plain"}, {"plain", "plain"}}

Combi[x_, y_] := Binomial[(x + y) - 1, y]
Combi[3, 2]
-> 6
Combi[10, 3]
->220

A better method therefore:

CombinWithRep[S_List, k_] := Module[{occupation, assignment},
  occupation =
   Flatten[Permutations /@
     IntegerPartitions[k, {Length[S]}, Range[0, k]], 1];
  assignment =
   Flatten[Table[ConstantArray[z, {#[[z]]}], {z, Length[#]}]] & /@
    occupation;
  Thread[S[[#]]] & /@ assignment
  ]

In[2]:= CombinWithRep[{"iced", "jam", "plain"}, 2]

Out[2]= {{"iced", "iced"}, {"jam", "jam"}, {"plain",
  "plain"}, {"iced", "jam"}, {"iced", "plain"}, {"jam", "plain"}}

Which can handle the Length[S] = 10, k=10 situation in still only seconds.

Mercury

comb.choose uses a nondeterministic list.member/2 to pick values from the list, and just puts them into a bag (a multiset). comb.choose_all gathers all of the possible bags that comb.choose would produce for a given list and number of picked values, and turns them into lists (for readability).

comb.count_choices shows off solutions.aggregate (which allows you to fold over solutions as they're found) rather than list.length and the factorial function.

:- module comb.
:- interface.
:- import_module list, int, bag.

:- pred choose(list(T)::in, int::in, bag(T)::out) is nondet.
:- pred choose_all(list(T)::in, int::in, list(list(T))::out) is det.
:- pred count_choices(list(T)::in, int::in, int::out) is det.

:- implementation.
:- import_module solutions.

choose(L, N, R) :- choose(L, N, bag.init, R).

:- pred choose(list(T)::in, int::in, bag(T)::in, bag(T)::out) is nondet.
choose(L, N, !R) :-
        ( N = 0 ->
                true
        ;
                member(X, L),
                bag.insert(!.R, X, !:R),
                choose(L, N - 1, !R)
        ).

choose_all(L, N, R) :-
        solutions(choose(L, N), R0),
        list.map(bag.to_list, R0, R).

count_choices(L, N, Count) :-
        aggregate(choose(L, N), count, 0, Count).

:- pred count(T::in, int::in, int::out) is det.
count(_, N0, N) :- N0 + 1 = N.

Usage:

:- module comb_ex.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module comb, list, string.

:- type doughtnuts
        --->    iced ; jam ; plain
        ;       glazed ; chocolate ; cream_filled ; mystery
        ;       cubed ; cream_covered ; explosive.

main(!IO) :-
        choose_all([iced, jam, plain], 2, L),
        count_choices([iced, jam, plain, glazed, chocolate, cream_filled,
                       mystery, cubed, cream_covered, explosive], 3, N),
        io.write(L, !IO), io.nl(!IO),
        io.write_string(from_int(N) ++ " choices.\n", !IO).

[[iced, iced], [jam, jam], [plain, plain], [iced, jam], [iced, plain], [jam, plain]]
220 choices.

Nim

import future, sequtils

proc combsReps[T](lst: seq[T], k: int): seq[seq[T]] =
  if k == 0:
    @[newSeq[T]()]
  elif lst.len == 0:
    @[]
  else:
    lst.combsReps(k - 1).map((x: seq[T]) => lst[0] & x) &
      lst[1 .. -1].combsReps(k)

echo(@["iced", "jam", "plain"].combsReps(2))
echo toSeq(1..10).combsReps(3).len

@[@[iced, iced], @[iced, jam], @[iced, plain], @[jam, jam], @[jam, plain], @[plain, plain]]
220

OCaml

let rec combs_with_rep k xxs =
  match k, xxs with
  | 0,  _ -> [[]]
  | _, [] -> []
  | k, x::xs ->
      List.map (fun ys -> x::ys) (combs_with_rep (k-1) xxs)
      @ combs_with_rep k xs

in the interactive loop:


# combs_with_rep 2 ["iced"; "jam"; "plain"] ;;
- : string list list =
[["iced"; "iced"]; ["iced"; "jam"]; ["iced"; "plain"]; ["jam"; "jam"];
 ["jam"; "plain"]; ["plain"; "plain"]]

# List.length (combs_with_rep 3 [1;2;3;4;5;6;7;8;9;10]) ;;
- : int = 220

Dynamic programming

let combs_with_rep m xs =
  let arr = Array.make (m+1) [] in
  arr.(0) <- [[]];
  List.iter (fun x ->
    for i = 1 to m do
      arr.(i) <- arr.(i) @ List.map (fun xs -> x::xs) arr.(i-1)
    done
  ) xs;
  arr.(m)

in the interactive loop:


# combs_with_rep 2 ["iced"; "jam"; "plain"] ;;
- : string list list =
[["iced"; "iced"]; ["jam"; "iced"]; ["jam"; "jam"]; ["plain"; "iced"];
 ["plain"; "jam"]; ["plain"; "plain"]]

# List.length (combs_with_rep 3 [1;2;3;4;5;6;7;8;9;10]) ;;
- : int = 220

PARI/GP

ways(k,v,s=[])={
	if(k==0,return([]));
	if(k==1,return(vector(#v,i,concat(s,[v[i]]))));
	if(#v==1,return(ways(k-1,v,concat(s,v))));
	my(u=vecextract(v,2^#v-2));
	concat(ways(k-1,v,concat(s,[v[1]])),ways(k,u,s))
};
xc(k,v)=binomial(#v+k-1,k);
ways(2, ["iced","jam","plain"])

Perl

The highly readable version:

sub p { $_[0] ? map p($_[0] - 1, [@{$_[1]}, $_[$_]], @_[$_ .. $#_]), 2 .. $#_ : $_[1] }
sub f { $_[0] ? $_[0] * f($_[0] - 1) : 1 }
sub pn{ f($_[0] + $_[1] - 1) / f($_[0]) / f($_[1] - 1) }

for (p(2, [], qw(iced jam plain))) {
        print "@$_\n";
}

printf "\nThere are %d ways to pick 7 out of 10\n", pn(7,10);

Prints:

iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

There are 11440 ways to pick 7 out of 10

With a module:

use Algorithm::Combinatorics qw/combinations_with_repetition/;
my $iter = combinations_with_repetition([qw/iced jam plain/], 2);
while (my $p = $iter->next) {
  print "@$p\n";
}
# Not an efficient way: generates them all in an array!
my $count =()= combinations_with_repetition([1..10],7);
print "There are $count ways to pick 7 out of 10\n";

Perl 6

One could simply generate all [[Permutations_with_repetitions#Perl_6|permutations]], and then remove "duplicates":

;
my $k = 2;

.put for [X](@S xx $k).unique(as => *.sort.cache, with => &[eqv])


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

Alternatively, a recursive solution:

proto combs_with_rep (UInt, @) {*}

multi combs_with_rep (0,  @)  { () }
multi combs_with_rep (1,  @a)  { map { $_, }, @a }
multi combs_with_rep ($,  []) { () }
multi combs_with_rep ($n, [$head, *@tail]) {
    |combs_with_rep($n - 1, ($head, |@tail)).map({ $head, |@_ }),
    |combs_with_rep($n, @tail);
}

.say for combs_with_rep( 2, [< iced jam plain >] );

# Extra credit:
sub postfix:<!> { [*] 1..$^n }
sub combs_with_rep_count ($k, $n) { ($n + $k - 1)! / $k! / ($n - 1)! }

say combs_with_rep_count( 3, 10 );

(iced iced)
(iced jam)
(iced plain)
(jam jam)
(jam plain)
(plain plain)
220

Phix

procedure choose(sequence from, integer n, at=1, sequence res={})
    if length(res)=n then
        ?res
    else
        for i=at to length(from) do
            choose(from,n,i,append(res,from[i]))
        end for
    end if
end procedure

choose({"iced","jam","plain"},2)


{"iced","iced"}
{"iced","jam"}
{"iced","plain"}
{"jam","jam"}
{"jam","plain"}
{"plain","plain"}

The second part suggests enough differences (collecting and showing vs only counting) to strike me as ugly and confusing. While I could easily, say, translate the C version, I'd rather forego the extra credit and use a completely different routine:

function choices(integer from, n, at=1, taken=0)
integer count = 0
    if taken=n then return 1 end if
    taken += 1
    for i=at to from do
        count += choices(from,n,i,taken)
    end for
    return count
end function

?choices(10,3)

PHP

Non-recursive algorithm to generate all combinations with repetitons. Taken from here: [https://habrahabr.ru/post/311934/] You must set k n variables and fill arrays b and c.


<?php
//Author Ivan Gavryushin @dcc0
$k=3;
$n=5;
//set amount of elements as in $n var
$c=array(1,2,3,4,5);
//set amount of 1 as in $k var
$b=array(1,1,1);
$j=$k-1;
//Вывод
	function printt($b,$k) {

	$z=0;

	while ($z < $k) print $b[$z++].' ';
	print '
';
}
printt ($b,$k);

        while (1) {
//Увеличение на позиции K до N

       	 if (array_search($b[$j]+1,$c)!==false ) {
  	      	$b[$j]=$b[$j]+1;
        	printt ($b,$k);
       }

       	if ($b[$k-1]==$n) {
	 $i=$k-1;
//Просмотр массива справа налево
	 while ($i >= 0) {
		//Условие выхода
	 	if ( $i == 0 && $b[$i] == $n) break 2;
//Поиск элемента для увеличения
       		  $m=array_search($b[$i]+1,$c);
		if ($m!==false) {
		           $c[$m]=$c[$m]-1;
			  $b[$i]=$b[$i]+1;

			$g=$i;
//Сортировка массива B
		while ($g != $k-1) {
			array_unshift ($c, $b[$g+1]);
			$b[$g+1]=$b[$i];
			$g++;
			}
//Удаление повторяющихся значений из C
	                            $c=array_diff($c,$b);
				    printt ($b,$k);
                                    array_unshift ($c, $n);

		 	     break;
       			 }
	 	$i--;

		}

	}


}

?>

PHP

<?php
  function combos($arr, $k) {
    if ($k == 0) {
      return array(array());
    }

    if (count($arr) == 0) {
      return array();
    }

    $head = $arr[0];

    $combos = array();
    $subcombos = combos($arr, $k-1);
    foreach ($subcombos as $subcombo) {
      array_unshift($subcombo, $head);
      $combos[] = $subcombo;
    }
    array_shift($arr);
    $combos = array_merge($combos, combos($arr, $k));
    return $combos;
  }

  $arr = array("iced", "jam", "plain");
  $result = combos($arr, 2);
  foreach($result as $combo) {
    echo implode(' ', $combo), "
";
  }
  $donuts = range(1, 10);
  $num_donut_combos = count(combos($donuts, 3));
  echo "$num_donut_combos ways to order 3 donuts given 10 types";
?>

{{out}} in the browser:


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
220 ways to order 3 donuts given 10 types

PicoLisp

(de combrep (N Lst)
   (cond
      ((=0 N) '(NIL))
      ((not Lst))
      (T
         (conc
            (mapcar
               '((X) (cons (car Lst) X))
               (combrep (dec N) Lst) )
            (combrep N (cdr Lst)) ) ) ) )

: (combrep 2 '(iced jam plain))
-> ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

: (length (combrep 3 (range 1 10)))
-> 220

PureBasic

Procedure nextCombination(Array combIndex(1), elementCount)
  ;combIndex() must be dimensioned to 'k' - 1, elementCount equals 'n' - 1
  ;combination produced includes repetition of elements and is represented by the array combIndex()
  Protected i, indexValue, combSize = ArraySize(combIndex()), curIndex

  ;update indexes
  curIndex = combSize
  Repeat
    combIndex(curIndex) + 1
    If combIndex(curIndex) > elementCount

      curIndex - 1
      If curIndex < 0
        For i = 0 To combSize
          combIndex(i) = 0
        Next
        ProcedureReturn #False ;array reset to first combination
      EndIf

    ElseIf curIndex < combSize

      indexValue = combIndex(curIndex)
      Repeat
        curIndex + 1
        combIndex(curIndex) = indexValue
      Until curIndex = combSize

    EndIf
  Until  curIndex = combSize

  ProcedureReturn #True ;array contains next combination
EndProcedure

Procedure.s display(Array combIndex(1), Array dougnut.s(1))
  Protected i, elementCount = ArraySize(combIndex()), output.s = "  "
  For i = 0 To elementCount
    output + dougnut(combIndex(i)) + " + "
  Next
  ProcedureReturn Left(output, Len(output) - 3)
EndProcedure

DataSection
  Data.s "iced", "jam", "plain"
EndDataSection

If OpenConsole()
  Define n = 3, k = 2, i, combinationCount
  Dim combIndex(k - 1)
  Dim dougnut.s(n - 1)
  For i = 0 To n - 1: Read.s dougnut(i): Next

  PrintN("Combinations of " + Str(k) + " dougnuts taken " + Str(n) + " at a time with repetitions.")
  combinationCount = 0
  Repeat
    PrintN(display(combIndex(), dougnut()))
    combinationCount + 1
  Until Not nextCombination(combIndex(), n - 1)
  PrintN("Total combination count: " + Str(combinationCount))

  ;extra credit
  n = 10: k = 3
  Dim combIndex(k - 1)
  combinationCount = 0
  Repeat: combinationCount + 1: Until Not nextCombination(combIndex(), n - 1)
  PrintN(#CRLF$ + "Ways to select " + Str(k) + " items from " + Str(n) + " types: " + Str(combinationCount))

  Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
  CloseConsole()
EndIf

The nextCombination() procedure operates on an array of indexes to produce the next combination. This generalization allows producing a combination from any collection of elements. nextCombination() returns the value #False when the indexes have reach their maximum values and are then reset.

Combinations of 2 dougnuts taken 3 at a time with repetitions.
  iced + iced
  iced + jam
  iced + plain
  jam + jam
  jam + plain
  plain + plain
Total combination count: 6

Ways to select 3 items from 10 types: 220

Python

 from itertools import combinations_with_replacement
>>> n, k = 'iced jam plain'.split(), 2
>>> list(combinations_with_replacement(n,k))
[('iced', 'iced'), ('iced', 'jam'), ('iced', 'plain'), ('jam', 'jam'), ('jam', 'plain'), ('plain', 'plain')]
>>> # Extra credit
>>> len(list(combinations_with_replacement(range(10), 3)))
220
>>>

'''References:'''

[http://docs.python.org/py3k/library/itertools.html#itertools.combinations_with_replacement Python documentation]

Or, assembling our own '''combsWithRep''', by composition of functional primitives:

'''Combinations with repetitions'''

from itertools import (accumulate, chain, islice, repeat)
from functools import (reduce)


# combsWithRep :: Int -> [a] -> [kTuple a]
def combsWithRep(k):
    '''A list of tuples, representing
       sets of cardinality k,
       with elements drawn from xs.
    '''
    def f(a, x):
        def go(ys, xs):
            return xs + [[x] + y for y in ys]
        return accumulate(a, go)

    def combsBySize(xs):
        return reduce(
            f, xs, chain(
                [[[]]],
                islice(repeat([]), k)
            )
        )
    return lambda xs: [
        tuple(x) for x in next(islice(
            combsBySize(xs), k, None
        ))
    ]


# TEST ----------------------------------------------------
def main():
    '''Test the generation of sets of cardinality
       k with elements drawn from xs.
    '''
    print(
        combsWithRep(2)(['iced', 'jam', 'plain'])
    )
    print(
        len(combsWithRep(3)(enumFromTo(0)(9)))
    )


# GENERIC -------------------------------------------------

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
    '''Integer enumeration from m to n.'''
    return lambda n: list(range(m, 1 + n))


# showLog :: a -> IO String
def showLog(*s):
    '''Arguments printed with
       intercalated arrows.'''
    print(
        ' -> '.join(map(str, s))
    )


# MAIN ---
if __name__ == '__main__':
    main()

[('iced', 'iced'), ('jam', 'iced'), ('jam', 'jam'), ('plain', 'iced'), ('plain', 'jam'), ('plain', 'plain')]
220

Racket


#lang racket
(define (combinations xs k)
  (cond [(= k 0)     '(())]
        [(empty? xs) '()]
        [(append (combinations (rest xs) k)
                 (map (λ(x) (cons (first xs) x))
                      (combinations xs (- k 1))))]))

REXX

version 1

This REXX version uses a type of anonymous recursion.

/*REXX pgm displays combination sets with repetitions for  X  things taken  Y  at a time*/
call RcombN    3,  2,  'iced jam plain'          /*The  1st  part of Rosetta Code task. */
call RcombN  -10,  3,  'Iced jam plain'          /* "   2nd    "   "    "      "    "   */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
RcombN: procedure; parse arg x,y,syms;  tell= x>0;  x=abs(x);   z=x+1  /*X>0? Show combo*/
        say copies('─',15) x "doughnut selection taken" y 'at a time:' /*display title. */
               do i=1  for words(syms);           $.i=word(syms, i)    /*assign symbols.*/
               end   /*i*/
        @.=1                                                           /*assign default.*/
               do #=1;        if tell    then  call show               /*display combos?*/
               @[email protected] + 1;   if @.y==z  then  if .(y-1)  then leave   /* ◄─── recursive*/
               end   /*#*/
        say copies('═',15)  #  "combinations.";    say;   say          /*display answer.*/
        return
/*──────────────────────────────────────────────────────────────────────────────────────*/
.: procedure expose @. y z;   parse arg ?;     if ?==0  then return 1;            p=@.? +1
        if p==z  then return .(? -1);      do j=?  to y;   @.j=p;   end  /*j*/;   return 0
/*──────────────────────────────────────────────────────────────────────────────────────*/
show:   L=;      do c=1  for y;   [email protected];   L=L $._;   end  /*c*/;       say L;    return


─────────────── 3 doughnut selection taken 2 at a time:
 iced iced
 iced jam
 iced plain
 jam jam
 jam plain
 plain plain
═══════════════ 6 combinations.


─────────────── 10 doughnut selection taken 3 at a time:
═══════════════ 220 combinations.

version 2

recursive (taken from version 1) Reformatted and variable names suitable for OoRexx.

/*REXX compute (and show) combination sets for nt things in ns places*/
  debug=0
  Call time 'R'
  Call RcombN 3,2,'iced,jam,plain'  /* The 1st part of the task      */
  Call RcombN -10,3,'iced,jam,plain,d,e,f,g,h,i,j' /* 2nd part       */
  Call RcombN -10,9,'iced,jam,plain,d,e,f,g,h,i,j' /* extra part     */
  Say time('E') 'seconds'
  Exit
/*-------------------------------------------------------------------*/
Rcombn: Procedure Expose thing. debug
  Parse Arg nt,ns,thinglist
  tell=nt>0
  nt=abs(nt)
  Say '------------' nt 'doughnut selection taken' ns 'at a time:'
  If tell=0 Then
    Say ' list output suppressed'
  Do i=1 By 1 While thinglist>''
    Parse Var thinglist thing.i ',' thinglist /* assign things.      */
    End
  index.=1
  Do cmb=1 By 1
    If tell Then                    /* display combinations          */
      Call show                     /* show this one                 */
    index.ns=index.ns+1
    Call show_index 'A'
    If index.ns==nt+1 Then
      If proc(ns-1) Then
        Leave
    End
  Say '------------' cmb 'combinations.'
  Say
  Return
/*-------------------------------------------------------------------*/
proc: Procedure Expose nt ns thing. index. debug
  Parse Arg recnt
  If recnt>0 Then Do
    p=index.recnt+1
    If p=nt+1 Then
      Return proc(recnt-1)
    Do i=recnt To ns
      index.i=p
      End
    Call show_index 'C'
    End
  Return recnt=0
/*-------------------------------------------------------------------*/
show: Procedure Expose index. thing. ns debug
  l=''
  Call show_index 'B----------------------->'
  Do i=1 To ns
    j=index.i
    l=l thing.j
    End
  Say l
  Return

show_index: Procedure Expose index. ns debug
  If debug Then Do
    Parse Arg tag
      l=tag
      Do i=1 To ns
        l=l index.i
        End
      Say l
    End
  Return

----------- 3 doughnut selection taken 2 at a time:
 iced iced
 iced jam
 iced plain
 jam jam
 jam plain
 plain plain
------------ 6 combinations.

------------ 10 doughnut selection taken 3 at a time:
 list output suppressed
------------ 220 combinations.

------------ 10 doughnut selection taken 9 at a time:
 list output suppressed
------------ 48620 combinations.

0.125000 seconds

version 3

iterative (transformed from version 1)

/*REXX compute (and show) combination sets for nt things in ns places*/
  Numeric Digits 20
  debug=0
  Call time 'R'
  Call IcombN 3,2,'iced,jam,plain'  /* The 1st part of the task      */
  Call IcombN -10,3,'iced,jam,plain,d,e,f,g,h,i,j' /* 2nd part       */
  Call IcombN -10,9,'iced,jam,plain,d,e,f,g,h,i,j' /* extra part     */
  Say time('E') 'seconds'
  Exit

IcombN: Procedure Expose thing. debug
  Parse Arg nt,ns,thinglist
  tell=nt>0
  nt=abs(nt)
  Say '------------' nt 'doughnut selection taken' ns 'at a time:'
  If tell=0 Then
    Say ' list output suppressed'
  Do i=1 By 1 While thinglist>''
    Parse Var thinglist thing.i ',' thinglist /* assign things.      */
    End
  index.=1
  cmb=0
  Call show
  i=ns+1
  Do While i>1
    i=i-1
    Do j=1 By 1 While index.i<nt
      index.i=index.i+1
      Call show
      End
    i1=i-1
    If index.i1<nt Then Do
      index.i1=index.i1+1
      Do ii=i To ns
        index.ii=index.i1
        End
      Call show
      i=ns+1
      End
    If index.1=nt Then Leave
    End
  Say cmb
  Return

show: Procedure Expose ns index. thing. tell cmb
  cmb=cmb+1
  If tell Then Do
    l=''
    Do i=1 To ns
      j=index.i
      l=l thing.j
      End
    Say l
    End
  Return

------------ 3 doughnut selection taken 2 at a time:
 iced iced
 iced jam
 iced plain
 jam jam
 jam plain
 plain plain
6
------------ 10 doughnut selection taken 3 at a time:
 list output suppressed
220
------------ 10 doughnut selection taken 9 at a time:
 list output suppressed
48620
0.109000 seconds

Slightly faster

Ring


# Project : Combinations with repetitions

n = 2
k = 3
temp = []
comb = []
num = com(n, k)
combinations(n, k)
comb = sortfirst(comb, 1)
see showarray(comb) + nl

func combinations(n, k)
while true
         temp = []
         for nr = 1 to k
               tm = random(n-1) + 1
               add(temp, tm)
         next
             add(comb, temp)
         for p = 1  to len(comb) - 1
               for q = p + 1 to len(comb)
                     if (comb[p][1] = comb[q][1]) and (comb[p][2] = comb[q][2]) and (comb[p][3] = comb[q][3])
                        del(comb, p)
                     ok
                next
         next
         if len(comb) = num
            exit
         ok
end

func com(n, k)
        res = pow(n, k)
        return res

func showarray(vect)
        svect = ""
        for nrs = 1 to len(vect)
              svect = "[" + vect[nrs][1] + " " + vect[nrs][2] + " " + vect[nrs][3] + "]" + nl
              see svect
        next

Func sortfirst(alist, ind)
        aList = sort(aList,ind)
        for n = 1 to len(alist)-1
             for m= n + 1 to len(aList)
                  if alist[n][1] = alist[m][1] and alist[m][2] < alist[n][2]
                     temp = alist[n]
                     alist[n] = alist[m]
                     alist[m] = temp
                   ok
             next
        next
        for n = 1 to len(alist)-1
             for m= n + 1 to len(aList)
                  if alist[n][1] = alist[m][1] and alist[n][2] = alist[m][2] and alist[m][3] < alist[n][3]
                     temp = alist[n]
                     alist[n] = alist[m]
                     alist[m] = temp
                   ok
             next
       next
       return aList

Output:


[1 1 1]
[1 1 2]
[1 2 1]
[1 2 2]
[2 1 1]
[2 1 2]
[2 2 1]
[2 2 2]

Ruby

possible_doughnuts = ['iced', 'jam', 'plain'].repeated_combination(2)
puts "There are #{possible_doughnuts.count} possible doughnuts:"
possible_doughnuts.each{|doughnut_combi| puts doughnut_combi.join(' and ')}

# Extra credit
possible_doughnuts = [*1..10].repeated_combination(3)
# size returns the size of the enumerator, or nil if it can’t be calculated lazily.
puts "", "#{possible_doughnuts.size} ways to order 3 donuts given 10 types."


There are 6 possible doughnuts:
iced and iced
iced and jam
iced and plain
jam and jam
jam and plain
plain and plain

220 ways to order 3 donuts given 10 types.

Scala

Scala has a combinations method in the standard library.


object CombinationsWithRepetition {

  def multi[A](as: List[A], k: Int): List[List[A]] =
    (List.fill(k)(as)).flatten.combinations(k).toList

  def main(args: Array[String]): Unit = {
    val doughnuts = multi(List("iced", "jam", "plain"), 2)
    for (combo <- doughnuts) println(combo.mkString(","))

    val bonus = multi(List(0,1,2,3,4,5,6,7,8,9), 3).size
    println("There are "+bonus+" ways to choose 3 items from 10 choices")
  }
}


iced,iced
iced,jam
iced,plain
jam,jam
jam,plain
plain,plain
There are 220 ways to choose 3 items from 10 choices

Scheme

(define (combs-with-rep k lst)
  (cond ((= k 0) '(()))
        ((null? lst) '())
        (else
         (append
          (map
           (lambda (x)
             (cons (car lst) x))
           (combs-with-rep (- k 1) lst))
          (combs-with-rep k (cdr lst))))))

(display (combs-with-rep 2 (list "iced" "jam" "plain"))) (newline)
(display (length (combs-with-rep 3 '(1 2 3 4 5 6 7 8 9 10)))) (newline)


((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))
220

Dynamic programming

(define (combs-with-rep m lst)
  (define arr (make-vector (+ m 1) '()))
  (vector-set! arr 0 '(()))
  (for-each (lambda (x)
	      (do ((i 1 (+ i 1)))
		  ((> i m))
		(vector-set! arr i (append (vector-ref arr i)
					   (map (lambda (xs) (cons x xs))
						(vector-ref arr (- i 1)))))
		)
	      ) lst)
  (vector-ref arr m))

(display (combs-with-rep 2 (list "iced" "jam" "plain"))) (newline)
(display (length (combs-with-rep 3 '(1 2 3 4 5 6 7 8 9 10)))) (newline)


((iced iced) (jam iced) (jam jam) (plain iced) (plain jam) (plain plain))
220

Sidef

func cwr (n, l, a = []) {
    n>0 ? (^l -> map {|k| __FUNC__(n-1, l.slice(k), [a..., l[k]]) }) : a
}

cwr(2, %w(iced jam plain)).each {|a|
    say a.map{ .join(' ') }.join("\n")
}

Also built-in:

%w(iced jam plain).combinations_with_repetition(2, {|*a|
    say a.join(' ')
})


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

Efficient count of the total number of combinations with repetition:

func cwr_count (n, m) { binomial(n + m - 1, m) }
printf("\nThere are %s ways to pick 7 out of 10 with repetition\n", cwr_count(10, 7))


There are 11440 ways to pick 7 out of 10 with repetition

Standard ML

let rec combs_with_rep k xxs =
  match k, xxs with
  | 0,  _ -> [[]]
  | _, [] -> []
  | k, x::xs ->
      List.map (fun ys -> x::ys) (combs_with_rep (k-1) xxs)
      @ combs_with_rep k xs

in the interactive loop:


- combs_with_rep (2, ["iced", "jam", "plain"]) ;
val it =
  [["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"],
   ["jam","plain"],["plain","plain"]] : string list list
- length (combs_with_rep (3, [1,2,3,4,5,6,7,8,9,10])) ;
val it = 220 : int

Dynamic programming

fun combs_with_rep (m, xs) = let
  val arr = Array.array (m+1, [])
in
  Array.update (arr, 0, [[]]);
  app (fn x =>
    Array.modifyi (fn (i, y) =>
      if i = 0 then y else y @ map (fn xs => x::xs) (Array.sub (arr, i-1))
    ) arr
  ) xs;
  Array.sub (arr, m)
end

in the interactive loop:


- combs_with_rep (2, ["iced", "jam", "plain"]) ;
val it =
  [["iced","iced"],["jam","iced"],["jam","jam"],["plain","iced"],
   ["plain","jam"],["plain","plain"]] : string list list
- length (combs_with_rep (3, [1,2,3,4,5,6,7,8,9,10])) ;
val it = 220 : int

Stata

function combrep(v,k) {
	n = cols(v)
	a = J(comb(n+k-1,k),k,v[1])
	u = J(1,k,1)
	for (i=2; 1; i++) {
		for (j=k; j>0; j--) {
			if (u[j]<n) break
		}
		if (j<1) return(a)
		m = u[j]+1
		for (; j<=k; j++) u[j] = m
		a[i,.] = v[u]
	}
}

combrep(("iced","jam","plain"),2)

a = combrep(1..10,3)
rows(a)

'''Output'''

           1       2
    +-----------------+
  1 |   iced    iced  |
  2 |   iced     jam  |
  3 |   iced   plain  |
  4 |    jam     jam  |
  5 |    jam   plain  |
  6 |  plain   plain  |
    +-----------------+

  220

Swift

(var objects: [T], n: Int) -> [[T]] {
  if n == 0 { return [[]] } else {
    var combos = [[T]]()
    while let element = objects.last {
      combos.appendContentsOf(combosWithRep(objects, n: n - 1).map{ $0 + [element] })
      objects.removeLast()
    }
    return combos
  }
}
print(combosWithRep(["iced", "jam", "plain"], n: 2).map {$0.joinWithSeparator(" and ")}.joinWithSeparator("\n"))

Output:

plain and plain
jam and plain
iced and plain
jam and jam
iced and jam
iced and iced

Tcl

package require Tcl 8.5
proc combrepl {set n {presorted no}} {
    if {!$presorted} {
        set set [lsort $set]
    }
    if {[incr n 0] < 1} {
	return {}
    } elseif {$n < 2} {
	return $set
    }
    # Recursive call
    set res [combrepl $set [incr n -1] yes]
    set result {}
    foreach item $set {
	foreach inner $res {
	    dict set result [lsort [list $item {*}$inner]] {}
	}
    }
    return [dict keys $result]
}

puts [combrepl {iced jam plain} 2]
puts [llength [combrepl {1 2 3 4 5 6 7 8 9 10} 3]]


{iced iced} {iced jam} {iced plain} {jam jam} {jam plain} {plain plain}
220

TXR

txr -p "(rcomb '(iced jam plain) 2)"


((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

txr -p "(length-list (rcomb '(0 1 2 3 4 5 6 7 8 9) 3))"

Ursala

#import std
#import nat

cwr = ~&s+ -<&*+ ~&K0=>&^|DlS/~& iota     # takes a set and a selection size

#cast %gLSnX

main = ^|(~&,length) cwr~~/(<'iced','jam','plain'>,2) ('1234567890',3)


(
   {
      <'iced','iced'>,
      <'iced','jam'>,
      <'iced','plain'>,
      <'jam','jam'>,
      <'jam','plain'>,
      <'plain','plain'>},
   220)

XPL0

code ChOut=8, CrLf=9, IntOut=11, Text=12;
int  Count, Array(10);

proc Combos(D, S, K, N, Names); \Generate all size K combinations of N objects
int  D, S, K, N, Names;         \depth of recursion, starting value of N, etc.
int  I;
[if D<K then                    \depth < size
    [for I:= S to N-1 do
        [Array(D):= I;
        Combos(D+1, I, K, N, Names);
        ];
    ]
else [Count:= Count+1;
     if Names(0) then
        [for I:= 0 to K-1 do
            [Text(0, Names(Array(I)));  ChOut(0, ^ )];
        CrLf(0);
        ];
     ];
];

[Count:= 0;
Combos(0, 0, 2, 3, ["iced", "jam", "plain"]);
Text(0, "Combos = ");  IntOut(0, Count);  CrLf(0);
Count:= 0;
Combos(0, 0, 3, 10, [0]);
Text(0, "Combos = ");  IntOut(0, Count);  CrLf(0);
]


iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
Combos = 6
Combos = 220

zkl

fcn combosK(k,seq){	// repeats, seq is finite
   if (k==1)    return(seq);
   if (not seq) return(T);
   self.fcn(k-1,seq).apply(T.extend.fp(seq[0])).extend(self.fcn(k,seq[1,*]));
}

combosK(2,T("iced","jam","plain")).apply("concat",",");
combosK(3,T(0,1,2,3,4,5,6,7,8,9)).len();


L("iced,iced","iced,jam","iced,plain","jam,jam","jam,plain","plain,plain")
220

ZX Spectrum Basic

10 READ n
20 DIM d$(n,5)
30 FOR i=1 TO n
40 READ d$(i)
50 NEXT i
60 DATA 3,"iced","jam","plain"
70 FOR i=1 TO n
80 FOR j=i TO n
90 PRINT d$(i);" ";d$(j)
100 NEXT j
110 NEXT i