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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 $\left\{\mathrm\left\{iced\right\}, \mathrm\left\{jam\right\}, \mathrm\left\{plain\right\}\right\}$, $|S| = 3$, 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''.

• 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]]

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

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));
end loop;
Set_IO.Put (Next);
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;


{{out}}

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


{{Out}}

{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

{{works with|BBC BASIC for Windows}}

      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%


{{out}}


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) );  {{out}} 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; 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; }  {{out}} 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 += 1; } } int main(){ vector<string> options{ "iced", "jam", "plain" }; combination_with_repetiton(3, 2, options); return 0; }  {{out}}  iced iced jam iced plain iced jam jam plain jam plain plain  ## C# {{trans|PHP}}  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; 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; } }  {{out}}  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 {{trans|Scheme}}  (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)))))  {{out}}  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].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"  {{out}}  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))))))  {{out}} ((:ICED :ICED) (:ICED :JAM) (:ICED :PLAIN) (:JAM :JAM) (:JAM :PLAIN) (:PLAIN :PLAIN))  ## Crystal {{trans|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 ")} # 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."  {{out}}  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); }  {{out}}  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 ~ L).array ~ combsRep(lst[1 ..$], k);
}

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


{{out}}

[["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"}))  {{out}}  {[|"iced" "iced"|] [|"iced" "jam"|] [|"jam" "jam"|] [|"iced" "plain"|] [|"jam" "plain"|] [|"plain" "plain"|]}  ## Elixir {{trans|Erlang}} 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)))}"  {{out}}  [: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).  {{out}}  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  {{out}}  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)) } 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"}))) }  {{out}}  [[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) }  {{out}}  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))) }  {{out}}  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]


{{out}}

[["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 $combsWithRep 3 [0 .. 9]  {{Out}} [["iced","iced"],["jam","iced"],["plain","iced"],["jam","jam"],["plain","jam"],["plain","plain"]] 220  =={{header|Icon}} and {{header|Unicon}}== Following procedure is a generator, which generates each combination of length n in turn:  # generate all combinations of length n from list L, # including repetitions procedure combinations_repetitions (L, n) if n = 0 then suspend [] # if reach 0, then return an empty list else if *L > 0 then { # keep the first element item := L # get all of length n in remaining list every suspend (combinations_repetitions (L[2:0], n)) # get all of length n-1 in remaining list # and add kept element to make list of size n every i := combinations_repetitions (L, n-1) do suspend [item] ||| i } end  Test procedure:  # convenience function procedure write_list (l) every (writes (!l || " ")) write () end # testing routine procedure main () # display all combinations for 2 of iced/jam/plain every write_list (combinations_repetitions(["iced", "jam", "plain"], 2)) # get a count for number of ways to select 3 items from 10 every push(num_list := [], 1 to 10) count := 0 every combinations_repetitions(num_list, 3) do count +:= 1 write ("There are " || count || " possible combinations of 3 from 10") end  {{out}}  plain plain jam plain jam jam iced plain iced jam iced iced There are 220 possible combinations of 3 from 10  =={{header|IS-BASIC}}== 100 PROGRAM "Combinat.bas" 110 READ N 120 STRING D$(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:  2 combr 3 0 0 0 1 0 2 1 1 1 2 2 2  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  {{out}}  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].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)) 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 ]; })();  {{Out}} [ [["iced", "iced"], ["iced", "jam"], ["iced", "plain"], ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]], 220 ]  ### ES6 {{Trans|Haskell}} (() => { '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 : 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))) }); })();  {{Out}} { "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."  {{Out}} $ 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|0.6}}

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))


{{out}}

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)
}


{{out}}


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))))
(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

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



This method will only work for small set and sample sizes (as it generates all Tuples then filters duplicates - Length[Tuples[Range,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;
]

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

Out= {{"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).


{{out}}

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


## Nim

{{trans|D}}

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 & x) &
lst[1 .. -1].combsReps(k)

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


{{out}}

@[@[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])),ways(k,u,s))
};
xc(k,v)=binomial(#v+k-1,k);
ways(2, ["iced","jam","plain"])


## Perl

sub p { $_ ? map p($_ - 1, [@{$_},$_[$_]], @_[$_ .. $#_]), 2 ..$#_ : $_ } sub f {$_ ? $_ * f($_ - 1) : 1 }
sub pn{ f($_ +$_ - 1) / f($_) / f($_ - 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 $count ways to pick 7 out of 10\n";  ## Perl 6 One could simply generate all [[Permutations_with_repetitions#Perl_6|permutations]], and then remove "duplicates": {{works with|Rakudo|2016.07}} ; my$k = 2;

.put for [X](@S xx $k).unique(as => *.sort.cache, with => &[eqv])  {{out}}  iced iced iced jam iced plain jam jam jam plain plain plain  Alternatively, a recursive solution: {{trans|Haskell}} proto combs_with_rep (UInt, @) {*} multi combs_with_rep (0, @) { () } multi combs_with_rep (1, @a) { map {$_, }, @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 );


{{out}}

(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)


{{out}}


{"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)


{{out}}


220



## 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 = $arr;$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)) ) ) ) )  {{out}} : (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.

{{out}}

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()


{{Out}}

[('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);           $.i=word(syms, i) /*assign symbols.*/ end /*i*/ @.=1 /*assign default.*/ do #=1; if tell then call show /*display combos?*/ @.y=@.y + 1; if @.y==z then if .(y-1) then leave /* ◄─── recursive*/ end /*#*/ say copies('═',15) # "combinations."; say; say /*display answer.*/ return /*──────────────────────────────────────────────────────────────────────────────────────*/ .: procedure expose @. y z; parse arg ?; if ?==0 then return 1; p=@.? +1 if p==z then return .(? -1); do j=? to y; @.j=p; end /*j*/; return 0 /*──────────────────────────────────────────────────────────────────────────────────────*/ show: L=; do c=1 for y; _=@.c; L=L$._;   end  /*c*/;       say L;    return


{{out|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.

/*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


{{out}}

----------- 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


{{out}}

------------ 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
next
for p = 1  to len(comb) - 1
for q = p + 1 to len(comb)
if (comb[p] = comb[q]) and (comb[p] = comb[q]) and (comb[p] = comb[q])
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] + " " + vect[nrs] + " " + vect[nrs] + "]" + 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] = alist[m] and alist[m] < alist[n]
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] = alist[m] and alist[n] = alist[m] and alist[m] < alist[n]
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

{{works with|Ruby|2.0}}

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."


{{out}}


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")
}
}



{{out}}


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


## Scheme

{{trans|PicoLisp}}

(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)


{{out}}


((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)


{{out}}


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



## Sidef

{{trans|Perl}}

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(' ')
})


{{out}}


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))


{{out}}


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)
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]]


{{out}}


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



## TXR

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


{{out}}


((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))"


{{out}}


220



## 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)


{{out}}


(
{
<'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, );
Text(0, "Combos = ");  IntOut(0, Count);  CrLf(0);
]


{{out}}


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



## zkl

{{trans|Clojure}}

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)).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();


{{out}}


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