Combinations with repetitions

Combinations with repetitions
You are encouraged to solve this task according to the task description, using any language you may know.

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

• k-combination with repetitions

See also

The number of samples of size k from n objects.

With   combinations and permutations   generation tasks.

	Order Unimportant	Order Important
Without replacement	${\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}$	$^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)$
	Task: Combinations	Task: Permutations
With replacement	${\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}$	$n^{k}$
	Task: Combinations with repetitions	Task: Permutations with repetitions

360 Assembly

<lang 360asm>* Combinations with repetitions - 16/04/2019 COMBREP CSECT

        USING  COMBREP,R13        base register
        B      72(R15)            skip savearea
        DC     17F'0'             savearea
        SAVE   (14,12)            save previous context
        ST     R13,4(R15)         link backward
        ST     R15,8(R13)         link forward
        LR     R13,R15            set addressability
        MVC    FLAVORS(9),SET1    flavors=3,draws=2,tell=1
        BAL    R14,COMBINE        call combine
        MVC    FLAVORS(9),SET2    flavors=10,draws=3,tell=0
        BAL    R14,COMBINE        call combine
        L      R13,4(0,R13)       restore previous savearea pointer
        RETURN (14,12),RC=0       restore registers from calling sav

COMBINE SR R9,R9 n=0

        MVI    V,X'00'            ~
        MVC    V+1(NN*L'V-1),V    v=0
      IF   CLI,TELL,EQ,X'01' THEN if tell then
        XPRNT  =C'list:',5          print
      ENDIF    ,                  endif

LOOP LA R6,1 i=1

      DO WHILE=(C,R6,LE,DRAWS)    do i=1 to draws
        LR     R1,R6                i
        SLA    R1,2                 ~
        L      R2,V-4(R1)           v(i)
        L      R3,FLAVORS           flavors
        BCTR   R3,0                 flavors-1
      IF    CR,R2,GT,R3 THEN        if v(i)>flavors-1 then
        LR     R1,R6                  i
        SLA    R1,2                   ~
        LA     R1,V(R1)               @v(i+1)
        L      R2,0(R1)               v(i+1)
        LA     R2,1(R2)               v(i+1)+1
        ST     R2,0(R1)               v(i+1)=v(i+1)+1
        LR     R7,R6                  j=i 
      DO WHILE=(C,R7,GE,=A(1))        do j=i to 1 by -1
        LR     R1,R6                    i
        SLA    R1,2                     ~
        L      R2,V(R1)                 v(i+1)
        LR     R1,R7                    j
        SLA    R1,2                     ~
        ST     R2,V-4(R1)               v(j)=v(i+1)
        BCTR   R7,0                     j--
      ENDDO    ,                      enddo j
      ENDIF    ,                    endif
        LA     R6,1(R6)             i++ 
      ENDDO    ,                  enddo i
        L      R1,DRAWS           draws
        LA     R1,1(R1)           draws+1
        SLA    R1,2               ~
        L      R2,V-4(R1)         v(draws+1)
        LTR    R2,R2              if v(draws+1)>0
        BP     EXITLOOP           then exit loop
        LA     R9,1(R9)           n=n+1
      IF   CLI,TELL,EQ,X'01' THEN if tell then
        MVC    BUF,=CL60' '         buf=' '
        LA     R10,1                ibuf=1
        LA     R6,1                 i=1 
      DO WHILE=(C,R6,LE,DRAWS)      do i=1 to draws
        LR     R1,R6                  i
        SLA    R1,2                   ~
        L      R1,V-4(R1)             v(i)
        MH     R1,=AL2(L'ITEMS)       ~
        LA     R4,ITEMS(R1)           @items(v(i)+1)
        LA     R5,BUF-1               @buf-1
        AR     R5,R10                 +ibuf
        MVC    0(6,R5),0(R4)          substr(buf,ibuf,6)=items(v(i)+1)
        LA     R10,L'ITEMS(R10)       ibuf=ibuf+length(items)
        LA     R6,1(R6)               i++ 
      ENDDO    ,                    enddo i
        XPRNT  BUF,L'BUF            print buf
      ENDIF    ,                    endif
        L      R2,V                 v(1)
        LA     R2,1(R2)             v(1)+1
        ST     R2,V                 v(1)=v(1)+1
        B      LOOP               loop

EXITLOOP L R1,FLAVORS flavors

        XDECO  R1,XDEC              edit flavors
        MVC    PG+4(2),XDEC+10      output flavors
        L      R1,DRAWS             draws 
        XDECO  R1,XDEC              edit draws
        MVC    PG+7(2),XDEC+10      output draws
        XDECO  R9,PG+11             edit & output n
        XPRNT  PG,L'PG              print buffer
        BR     R14                return

NN EQU 16 ITEMS DC CL6'iced',CL6'jam',CL6'plain' FLAVORS DS F DRAWS DS F TELL DS X SET1 DC F'3',F'2',X'01' flavors=3,draws=2,tell=1 SET2 DC F'10',F'3',X'00' flavors=10,draws=3,tell=0 V DS (NN)F v(nn) BUF DS CL60 buf PG DC CL40'cwr(..,..)=............' XDEC DS CL12 temp for xdeco

        REGEQU
        END    COMBREP </lang>

Output:

list:
iced  iced
jam   iced
plain iced
jam   jam
plain jam
plain plain
cwr( 3, 2)=           6
cwr(10, 3)=         220

Ada

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

combinations.adb: <lang Ada>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;</lang>

Output:

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

AppleScript

Translation of: Haskell

Translation of: Python

<lang 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</lang>

Output:

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

AutoHotkey

<lang AutoHotkey>;===========================================================

based on "https://www.geeksforgeeks.org/combinations-with-repetitions/"

===========================================================

CombinationRepetition(arr, k:=0, Delim:="") { CombinationRepetitionUtil(arr, k?k:str.count(), Delim, [k+1], result:=[]) return result } ;=========================================================== CombinationRepetitionUtil(arr, k, Delim, chosen, result , index:=1, start:=1){ line := [], i:=0, res := "" if (index = k+1){ while (++i <= k) res .= arr[chosen[i]] Delim, line.push(arr[chosen[i]]) return result.Push(Trim(res, Delim)) } i:=start while (i <= arr.count()) chosen[Index]:=i, CombinationRepetitionUtil(arr, k, Delim, chosen, result, index+1, i++) } ;===========================================================</lang> Examples:<lang AutoHotkey>result := CombinationRepetition(["iced","jam","plain"], 2, " + ") for k, v in result res .= v "`n" res := trim(res, ",") "`n" MsgBox % result.count() " Combinations with Repetition found:`n" res MsgBox % CombinationRepetition([0,1,2,3,4,5,6,7,8,9], 3).Count()</lang>

Outputs:

---------------------------
6 Combinations with Repetition found:
iced + iced
iced + jam
iced + plain
jam + jam
jam + plain
plain + plain
---------------------------
220
---------------------------

AWK

<lang AWK>

1. 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)

} </lang>

output:

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

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> 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%</lang>

Output:

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. <lang bracmat>( ( 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) );</lang>

Output:

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

C

<lang 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;

} </lang>

Output:

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#

Translation of: PHP

<lang csharp> 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;
   }

} </lang>

Output:

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

Recursive version <lang csharp> 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));
   }
 }

}

</lang>

C++

Non recursive version. <lang cpp>

1. include <cstdio>
2. include <vector>
3. 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; 

} </lang>

Output:

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

Clojure

Translation of: Scheme

<lang 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)))))

</lang>

Output:

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

CoffeeScript

<lang 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" </lang>

Output:

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. <lang lisp>(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)))))) </lang>

Output:

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

Crystal

Translation of: Ruby

<lang ruby>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 ")}

1. Extra credit

possible_doughnuts = (1..10).to_a.repeated_combinations(3)

1. 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."</lang>

Output:

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 lexicographic next bit permutation to generate combinations with repetitions. <lang d>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);

}</lang>

Output:

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

Short Recursive Version

<lang d>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;

}</lang>

Output:

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

EasyLang

<lang>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"</lang>

Output:

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. <lang scheme>

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

</lang>

Egison

<lang 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"})) </lang>

Output:

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

Elixir

Translation of: Erlang

<lang 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)))}"</lang>

Output:

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

Extra credit: 220

Erlang

<lang 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).

</lang>

Output:

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

<lang 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 </lang>

Output:

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

GAP

<lang gap># Built-in UnorderedTuples(["iced", "jam", "plain"], 2);</lang>

Go

Concise recursive

<lang go>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"})))

}</lang>

Output:

[[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. <lang go>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) }</lang>

Output:

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. <lang go>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)))

}</lang>

Output:

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

<lang 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]</lang>

Output:

[["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:

<lang haskell>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"]</lang>

and another approach, using manual recursion: <lang haskell>combsWithRep

 :: (Eq a)
 => Int -> [a] -> a

combsWithRep k xs = comb k []

 where
   comb 0 ys = ys
   comb n [] = comb (pred n) (pure <$> xs)
   comb n peers = comb (pred n) (peers >>= nextLayer)
     where
       nextLayer ys@(h:_) = (: ys) <$> dropWhile (/= h) xs

main :: IO () main = do

 print $ combsWithRep 2 ["iced", "jam", "plain"]
 print $ length $ combsWithRep 3 [0 .. 9]</lang>

Output:

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

Icon and Unicon

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

1. generate all combinations of length n from list L,
2. 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 </lang>

Test procedure:

<lang Icon>

1. convenience function

procedure write_list (l)

 every (writes (!l || " "))
 write ()

end

1. 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 </lang>

Output:

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

IS-BASIC

<lang 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</lang>

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.

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

Example use:

<lang j> 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</lang>

J Alternate implementation

Considerably faster:

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

rcomb functions identically, and combr calculates indices:

<lang j> 2 combr 3 0 0 0 1 0 2 1 1 1 2 2 2</lang>

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 <lang 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 </lang>

MultiCombinations.java <lang 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 </lang>

Output:

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

<lang javascript><html><head><title>Donuts</title></head>

<body>

<script type="application/javascript">

function disp(x) { var e = document.createTextNode(x + '\n'); document.getElementById('x').appendChild(e); }

function pick(n, got, pos, from, show) { var cnt = 0; if (got.length == n) { if (show) disp(got.join(' ')); return 1; } for (var i = pos; i < from.length; i++) { got.push(from[i]); cnt += pick(n, got, i, from, show); got.pop(); } return cnt; }

disp(pick(2, [], 0, ["iced", "jam", "plain"], true) + " combos"); disp("pick 3 out of 10: " + pick(3, [], 0, "a123456789".split(), false) + " combos"); </script></body></html></lang>

Output:

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

Functional

<lang JavaScript>(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
   ];

})();</lang>

Output:

<lang JavaScript>[

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

]</lang>

ES6

Translation of: Haskell

<lang JavaScript>(() => {

   '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)))
   });

})();</lang>

Output:

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

jq

<lang 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) ;</lang>

The task: <lang jq> "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."

</lang>

Output:

$ 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

Works with: Julia version 0.6

<lang 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))</lang>

Output:

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

<lang scala>// 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".chunked(1)
   CombsWithReps(3, 10, generic10, true)   

}</lang>

Output:

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

Translation of: Erlang

and

Translation of: Clojure

With List Comprehension

<lang lisp> (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))))

</lang>

With Map

<lang lisp> (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))))

</lang>

Output is the same for both:

<lang lisp> > (combinations '(iced jam plain) 2) ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) </lang>

Lobster

Translation of: C

<lang Lobster>import std

// set S of length n, choose k

def choose(s, k, f):

   let got = map(k): s[0]
   let n = s.length

   def choosi(n_chosen, at):
       var count = 0
       if n_chosen == k:
           f(got)
           return 1
       var i = at
       while i < n:
           got[n_chosen] = s[i]
           count += choosi(n_chosen + 1, i)
           i += 1
       return count
   return choosi(0, 0)

let count = choose(["iced", "jam", "plain"], 2): print(_) print count let extra = choose(map(10):_, 3): _ print extra</lang>

Output:

["iced", "iced"]
["iced", "jam"]
["iced", "plain"]
["jam", "jam"]
["jam", "plain"]
["plain", "plain"]
6
220

Lua

<lang 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 </lang>

Mathematica / 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). <lang Mathematica>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 </lang>

A better method therefore: <lang Mathematica>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"}}

</lang>

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.

<lang Mercury>:- 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.</lang>

Usage:

<lang Mercury>:- 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).</lang>

Output:

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

Nim

Translation of: D

<lang 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</lang>

Output:

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

OCaml

Translation of: Haskell

<lang 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</lang>

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

<lang ocaml>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)</lang>

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

<lang parigp>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"])</lang>

Perl

The highly readable version: <lang perl>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); </lang>

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: <lang perl>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";

}

1. 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";</lang>

Phix

<lang 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)</lang>

Output:

{"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: <lang Phix>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)</lang>

Output:

220

PHP

<lang 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";

?></lang>

Output:

in the browser

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

Non-recursive algorithm to generate all combinations with repetitons. Taken from here: [1]

You must set k n variables and fill arrays b and c. 

<lang PHP> <?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--;

}

} }

?> </lang>

PicoLisp

<lang 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)) ) ) ) )</lang>

Output:

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

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

Prolog

<lang prolog> combinations_of_length(_,[]). combinations_of_length([X|T],[X|Combinations]):-

   combinations_of_length([X|T],Combinations).

combinations_of_length([_|T],[X|Combinations]):-

   combinations_of_length(T,[X|Combinations]).

</lang>

?- [user].
|: combinations_of_length(_,[]).
|: combinations_of_length([X|T],[X|Combinations]):-
|:     combinations_of_length([X|T],Combinations).
|: combinations_of_length([_|T],[X|Combinations]):-
|:     combinations_of_length(T,[X|Combinations]).
|: ^D% user://1 compiled 0.01 sec, 3 clauses
true.

?- combinations_of_length([0,1,2,4],[L0, L1, L2, L3, L4, L5]).
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = L5, L5 = 0 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 2,
.
.
.

PureBasic

<lang 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(n) + " dougnut types taken " + Str(k) + " 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 </lang>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.

Output:

Combinations of 3 dougnut types taken 2 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

<lang 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 >>> </lang>

References:

• Python documentation

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

Translation of: Haskell

Works with: Python version 3.7

<lang python>Combinations with repetitions

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

1. 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
       ))
   ]

1. 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)))
   )

1. GENERIC -------------------------------------------------

1. enumFromTo :: (Int, Int) -> [Int]

def enumFromTo(m):

   Integer enumeration from m to n.
   return lambda n: list(range(m, 1 + n))

1. showLog :: a -> IO String

def showLog(*s):

   Arguments printed with
      intercalated arrows.
   print(
       ' -> '.join(map(str, s))
   )

1. MAIN ---

if __name__ == '__main__':

   main()</lang>

Output:

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

Racket

<lang racket>

1. 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))))]))

</lang>

Raku

(formerly Perl 6)

One could simply generate all permutations, and then remove "duplicates":

Works with: Rakudo version 2016.07

<lang perl6>my @S = <iced jam plain>; my $k = 2;

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

Output:

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

Alternatively, a recursive solution:

Translation of: Haskell

<lang perl6>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 >] );

1. 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 );</lang>

Output:

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

REXX

version 1

This REXX version uses a type of anonymous recursion. <lang rexx>/*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?*/
              @.y=@.y + 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; _=@.c; L=L $._; end /*c*/; say L; return</lang>

output:

─────────────── 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. <lang rexx>/*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</lang>

Output:

----------- 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) <lang rexx>/*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</lang>

Output:

------------ 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

<lang ring>

1. 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

</lang> 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

Works with: Ruby version 2.0

<lang 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 ')}

1. Extra credit

possible_doughnuts = [*1..10].repeated_combination(3)

1. 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."</lang>

Output:

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. <lang scala> 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")
 }

} </lang>

Output:

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

Scheme

Translation of: PicoLisp

<lang 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)</lang>

Output:

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

Dynamic programming

<lang scheme>(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)</lang>

Output:

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

Sidef

Translation of: Perl

<lang ruby>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")

}</lang>

Also built-in:

<lang ruby>%w(iced jam plain).combinations_with_repetition(2, {|*a|

   say a.join(' ')

})</lang>

Output:

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

Efficient count of the total number of combinations with repetition: <lang ruby>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))</lang>

Output:

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

Standard ML

Translation of: Haskell

<lang sml>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</lang>

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

<lang sml>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</lang>

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

<lang 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)</lang>

Output

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

  220

Swift

<lang Swift>func combosWithRep<T>(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"))</lang> Output:

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

Tcl

<lang 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]]</lang>

Output:

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

TXR

<lang dos>txr -p "(rcomb '(iced jam plain) 2)"</lang>

Output:

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

---

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

Output:

220

Ursala

<lang Ursala>#import std

1. import nat

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

1. cast %gLSnX

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

Output:

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

VBScript

<lang vb>' Combinations with repetitions - iterative - VBScript

Sub printc(vi,n,vs) Dim i, w For i=0 To n-1 w=w &" "& vs(vi(i)) Next 'i Wscript.Echo w End Sub

Sub combine(flavors, draws, xitem, tell) Dim n, i, j ReDim v(draws) If tell Then Wscript.Echo "list of cwr("& flavors &","& draws &") :" Do While True For i=0 To draws-1 If v(i)>flavors-1 Then v(i+1)=v(i+1)+1 For j=i To 0 Step -1 v(j)=v(i+1) Next 'j End If Next 'i If v(draws)>0 Then Exit Do n=n+1 If tell Then Call printc(v, draws, xitem) v(0)=v(0)+1 Loop Wscript.Echo "cwr("& flavors &","& draws &")="&n End Sub

items=Array( "iced", "jam", "plain" ) combine 3, 2, items, True combine 10, 3, , False combine 10, 7, , False combine 10, 9, , False </lang>

Output:

list of cwr(3,2) :
 iced iced
 jam iced
 plain iced
 jam jam
 plain jam
 plain plain
cwr(3,2)=6
cwr(10,3)=220
cwr(10,7)=11440
cwr(10,9)=48620

Wren

Translation of: Go

<lang ecmascript>var combrep // recursive 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)</lang>

Output:

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

XPL0

<lang 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); ]</lang>

Output:

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

zkl

Translation of: Clojure

<lang 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,*]));

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

Output:

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

ZX Spectrum Basic

<lang zxbasic>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</lang>