# Combinations

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

Given non-negative integers   m   and   n,   generate all size   m   combinations   of the integers from   0   (zero)   to   n-1   in sorted order   (each combination is sorted and the entire table is sorted).

Example

3   comb   5     is:

0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4


If it is more "natural" in your language to start counting from   1   (unity) instead of   0   (zero),
the combinations can be of the integers from   1   to   n.

The number of samples of size k from n objects.

With   combinations and permutations   generation tasks.

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

## 360 Assembly

Translation of: C

Nice algorithm without recursion borrowed from C. Recursion is elegant but iteration is efficient. For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. <lang 360asm>* Combinations 26/05/2016 COMBINE CSECT

        USING  COMBINE,R13        base register
B      72(R15)            skip savearea
DC     17F'0'             savearea
STM    R14,R12,12(R13)    prolog
ST     R13,4(R15)         "
ST     R15,8(R13)         "
LR     R13,R15            "
SR     R3,R3              clear
LA     R7,C               @c(1)
LH     R8,N               v=n


LOOPI1 STC R8,0(R7) do i=1 to n; c(i)=n-i+1

        LA     R7,1(R7)             @c(i)++
BCT    R8,LOOPI1          next i


LOOPBIG LA R10,PG big loop {------------------

        LH     R1,N               n
LA     R7,C-1(R1)         @c(i)
LH     R6,N               i=n


LOOPI2 IC R3,0(R7) do i=n to 1 by -1; r2=c(i)

        XDECO  R3,PG+80             edit c(i)
MVC    0(2,R10),PG+90       output c(i)
LA     R10,3(R10)           @pgi=@pgi+3
BCTR   R7,0                 @c(i)--
BCT    R6,LOOPI2          next i
XPRNT  PG,80              print buffer
LA     R7,C               @c(1)
LH     R8,M               v=m
LA     R6,1               i=1


LOOPI3 LR R1,R6 do i=1 by 1; r1=i

        IC     R3,0(R7)             c(i)
CR     R3,R8                while c(i)>=m-i+1
BL     ELOOPI3              leave i
CH     R6,N                 if i>=n
BNL    ELOOPBIG             exit loop
BCTR   R8,0                 v=v-1
LA     R7,1(R7)             @c(i)++
LA     R6,1(R6)             i=i+1
B      LOOPI3             next i


ELOOPI3 LR R1,R6 i

        LA     R4,C-1(R1)         @c(i)
IC     R3,0(R4)           c(i)
LA     R3,1(R3)           c(i)+1
STC    R3,0(R4)           c(i)=c(i)+1
BCTR   R7,0               @c(i)--


LOOPI4 CH R6,=H'2' do i=i to 2 by -1

        BL     ELOOPI4            leave i
IC     R3,1(R7)             c(i)
LA     R3,1(R3)             c(i)+1
STC    R3,0(R7)             c(i-1)=c(i)+1
BCTR   R7,0                 @c(i)--
BCTR   R6,0                 i=i-1
B      LOOPI4             next i


ELOOPI4 B LOOPBIG big loop }------------------ ELOOPBIG L R13,4(0,R13) epilog

        LM     R14,R12,12(R13)    "
XR     R15,R15            "
BR     R14                exit


M DC H'5' <=input N DC H'3' <=input C DS 64X array of 8 bit integers PG DC CL92' ' buffer

        YREGS
END    COMBINE</lang>

Output:
 1  2  3
1  2  4
1  2  5
1  3  4
1  3  5
1  4  5
2  3  4
2  3  5
2  4  5
3  4  5


procedure Test_Combinations is

  generic
type Integers is range <>;
package Combinations is
type Combination is array (Positive range <>) of Integers;
procedure First (X : in out Combination);
procedure Next (X : in out Combination);
procedure Put (X : Combination);
end Combinations;

package body Combinations is
procedure First (X : in out Combination) is
begin
X (1) := Integers'First;
for I in 2..X'Last loop
X (I) := X (I - 1) + 1;
end loop;
end First;
procedure Next (X : in out Combination) is
begin
for I in reverse X'Range loop
if X (I) < Integers'Val (Integers'Pos (Integers'Last) - X'Last + I) then
X (I) := X (I) + 1;
for J in I + 1..X'Last loop
X (J) := X (J - 1) + 1;
end loop;
return;
end if;
end loop;
raise Constraint_Error;
end Next;
procedure Put (X : Combination) is
begin
for I in X'Range loop
Put (Integers'Image (X (I)));
end loop;
end Put;
end Combinations;

type Five is range 0..4;
package Fives is new Combinations (Five);
use Fives;

  X : Combination (1..3);


begin

  First (X);
loop
Put (X); New_Line;
Next (X);
end loop;


exception

  when Constraint_Error =>
null;


end Test_Combinations;</lang> The solution is generic the formal parameter is the integer type to make combinations of. The type range determines n. In the example it is <lang ada>type Five is range 0..4;</lang> The parameter m is the object's constraint. When n < m the procedure First (selects the first combination) will propagate Constraint_Error. The procedure Next selects the next combination. Constraint_Error is propagated when it is the last one.

Output:
 0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4


## ALGOL 68

Translation of: Python
Works with: ALGOL 68 version Revision 1 - one minor extension to language used - PRAGMA READ, similar to C's #include directive.
Works with: ALGOL 68G version Any - tested with release algol68g-2.6.

File: prelude_combinations.a68<lang algol68># -*- coding: utf-8 -*- #

COMMENT REQUIRED BY "prelude_combinations_generative.a68"

 MODE COMBDATA = ~;


PROVIDES:

1. COMBDATA*=~* #
2. comb*=~ list* #

END COMMENT

MODE COMBDATALIST = REF[]COMBDATA; MODE COMBDATALISTYIELD = PROC(COMBDATALIST)VOID;

PROC comb gen combinations = (INT m, COMBDATALIST list, COMBDATALISTYIELD yield)VOID:(

 CASE m IN
# case 1: transpose list #
FOR i TO UPB list DO yield(list[i]) OD
OUT
[m + LWB list - 1]COMBDATA out;
INT index out := 1;
FOR i TO UPB list DO
COMBDATA first = list[i];
# FOR COMBDATALIST sub recombination IN # comb gen combinations(m - 1, list[i+1:] #) DO (#,
##   (COMBDATALIST sub recombination)VOID:(
out[LWB list   ] := first;
out[LWB list+1:] := sub recombination;
yield(out)
# OD #))
OD
ESAC


);

SKIP</lang>File: test_combinations.a68<lang algol68>#!/usr/bin/a68g --script #

1. -*- coding: utf-8 -*- #

CO REQUIRED BY "prelude_combinations.a68" CO

 MODE COMBDATA = INT;

1. PROVIDES:#
2. COMBDATA~=INT~ #
3. comb ~=int list ~#

FORMAT data fmt = $g(0)$;

main:(

 INT m = 3;
FORMAT list fmt = $"("n(m-1)(f(data fmt)",")f(data fmt)")"$;
FLEX[0]COMBDATA test data list := (1,2,3,4,5);

1. FOR COMBDATALIST recombination data IN # comb gen combinations(m, test data list #) DO (#,
1. (COMBDATALIST recombination)VOID:(
   printf ((list fmt, recombination, $l$))

1. OD # ))

) </lang>

Output:
(1,2,3)
(1,2,4)
(1,2,5)
(1,3,4)
(1,3,5)
(1,4,5)
(2,3,4)
(2,3,5)
(2,4,5)
(3,4,5)


## AppleScript

### Iteration

<lang AppleScript>on comb(n, k)

   set c to {}
repeat with i from 1 to k
set end of c to i's contents
end repeat
set r to {c's contents}
repeat while my next_comb(c, k, n)
set end of r to c's contents
end repeat
return r


end comb

on next_comb(c, k, n)

   set i to k
set c's item i to (c's item i) + 1
repeat while (i > 1 and c's item i ≥ n - k + 1 + i)
set i to i - 1
set c's item i to (c's item i) + 1
end repeat
if (c's item 1 > n - k + 1) then return false
repeat with i from i + 1 to k
set c's item i to (c's item (i - 1)) + 1
end repeat
return true


end next_comb

return comb(5, 3)</lang>

Output:

<lang AppleScript>{{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}}</lang>

### Functional composition

Translation of: JavaScript

<lang AppleScript>-- comb :: Int -> [a] -> a on comb(n, lst)

   if n < 1 then
{{}}
else
if not isNull(lst) then
set {h, xs} to uncons(lst)

map(curry(my cons)'s |λ|(h), comb(n - 1, xs)) & comb(n, xs)
else
{}
end if
end if


end comb

-- TEST ----------------------------------------------------------------------- on run

   intercalate(linefeed, ¬
map(unwords, comb(3, enumFromTo(0, 4))))



end run

-- GENERIC FUNCTIONS ----------------------------------------------------------

-- cons :: a -> [a] -> [a] on cons(x, xs)

   {x} & xs


end cons

-- curry :: (Script|Handler) -> Script on curry(f)

   script
on |λ|(a)
script
on |λ|(b)
|λ|(a, b) of mReturn(f)
end |λ|
end script
end |λ|
end script


end curry

-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)

   if n < m then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst


end enumFromTo

-- intercalate :: Text -> [Text] -> Text on intercalate(strText, lstText)

   set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined


end intercalate

-- isNull :: [a] -> Bool on isNull(xs)

   if class of xs is string then
xs = ""
else
xs = {}
end if


end isNull

-- 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

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)

   if class of f is script then
f
else
script
property |λ| : f
end script
end if


end mReturn

-- uncons :: [a] -> Maybe (a, [a]) on uncons(xs)

   set lng to length of xs
if lng > 0 then
if class of xs is string then
set cs to text items of xs
{item 1 of cs, rest of cs}
else
{item 1 of xs, rest of xs}
end if
else
missing value
end if


end uncons

-- unwords :: [String] -> String on unwords(xs)

   intercalate(space, xs)


end unwords</lang>

Output:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4

## AutoHotkey

contributed by Laszlo on the ahk forum <lang AutoHotkey>MsgBox % Comb(1,1) MsgBox % Comb(3,3) MsgBox % Comb(3,2) MsgBox % Comb(2,3) MsgBox % Comb(5,3)

Comb(n,t) { ; Generate all n choose t combinations of 1..n, lexicographically

  IfLess n,%t%, Return
Loop %t%
c%A_Index% := A_Index
i := t+1, c%i% := n+1

  Loop {
Loop %t%
i := t+1-A_Index, c .= c%i% " "
c .= "n"     ; combinations in new lines
j := 1, i := 2
Loop
If (c%j%+1 = c%i%)
c%j% := j, ++j, ++i
Else Break
If (j > t)
Return c
c%j% += 1
}


}</lang>

## AWK

<lang awk>BEGIN { ## Default values for r and n (Choose 3 from pool of 5). Can ## alternatively be set on the command line:- ## awk -v r=<number of items being chosen> -v n=<how many to choose from> -f <scriptname> if (length(r) == 0) r = 3 if (length(n) == 0) n = 5

for (i=1; i <= r; i++) { ## First combination of items: A[i] = i if (i < r ) printf i OFS else print i}

## While 1st item is less than its maximum permitted value... while (A[1] < n - r + 1) { ## loop backwards through all items in the previous ## combination of items until an item is found that is ## less than its maximum permitted value: for (i = r; i >= 1; i--) { ## If the equivalently positioned item in the ## previous combination of items is less than its ## maximum permitted value... if (A[i] < n - r + i) { ## increment the current item by 1: A[i]++ ## Save the current position-index for use ## outside this "for" loop: p = i break}} ## Put consecutive numbers in the remainder of the array, ## counting up from position-index p. for (i = p + 1; i <= r; i++) A[i] = A[i - 1] + 1

## Print the current combination of items: for (i=1; i <= r; i++) { if (i < r) printf A[i] OFS else print A[i]}} exit}</lang>

Usage:

awk -v r=3 -v n=5 -f combn.awk

Output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5


## BBC BASIC

<lang bbcbasic> INSTALL @lib$+"SORTLIB"  sort% = FN_sortinit(0,0) M% = 3 N% = 5 C% = FNfact(N%)/(FNfact(M%)*FNfact(N%-M%)) DIM s$(C%)
PROCcomb(M%, N%, s$()) CALL sort%, s$(0)
FOR I% = 0 TO C%-1
PRINT s$(I%) NEXT END DEF PROCcomb(C%, N%, s$())
LOCAL I%, U%
FOR U% = 0 TO 2^N%-1
IF FNbits(U%) = C% THEN
s$(I%) = FNlist(U%) I% += 1 ENDIF NEXT ENDPROC DEF FNbits(U%) LOCAL N% WHILE U% N% += 1 U% = U% AND (U%-1) ENDWHILE = N% DEF FNlist(U%) LOCAL N%, s$
WHILE U%
IF U% AND 1 s$+= STR$(N%) + " "
N% += 1
U% = U% >> 1
ENDWHILE
= s$DEF FNfact(N%) IF N%<=1 THEN = 1 ELSE = N%*FNfact(N%-1)  </lang> ## Bracmat The program first constructs a pattern with m variables and an expression that evaluates m variables into a combination. Then the program constructs a list of the integers 0 ... n-1. The real work is done in the expression !list:!pat. When a combination is found, it is added to the list of combinations. Then we force the program to backtrack and find the next combination by evaluating the always failing ~. When all combinations are found, the pattern fails and we are in the rhs of the last | operator. <lang bracmat>(comb=  bvar combination combinations list m n pat pvar var  . !arg:(?m.?n)  & ( pat = ? & !combinations (.!combination):?combinations & ~ ) & :?list:?combination:?combinations & whl ' ( !m+-1:~<0:?m & chu$(utf$a+!m):?var & glf$('(%@?.$var)):(=?pvar) & '(? ()$pvar ()$pat):(=?pat) & glf$('(!.$var)):(=?bvar) & ( '$combination:(=)
& '$bvar:(=?combination) | '($bvar ()$combination):(=?combination) ) ) & whl ' (!n+-1:~<0:?n&!n !list:?list) & !list:!pat | !combinations);</lang> comb$(3.5)

(.0 1 2)
(.0 1 3)
(.0 1 4)
(.0 2 3)
(.0 2 4)
(.0 3 4)
(.1 2 3)
(.1 2 4)
(.1 3 4)
(.2 3 4)


## C

<lang C>#include <stdio.h>

/* Type marker stick: using bits to indicate what's chosen. The stick can't

* handle more than 32 items, but the idea is there; at worst, use array instead */


typedef unsigned long marker; marker one = 1;

void comb(int pool, int need, marker chosen, int at) { if (pool < need + at) return; /* not enough bits left */

if (!need) { /* got all we needed; print the thing. if other actions are * desired, we could have passed in a callback function. */ for (at = 0; at < pool; at++) if (chosen & (one << at)) printf("%d ", at); printf("\n"); return; } /* if we choose the current item, "or" (|) the bit to mark it so. */ comb(pool, need - 1, chosen | (one << at), at + 1); comb(pool, need, chosen, at + 1); /* or don't choose it, go to next */ }

int main() { comb(5, 3, 0, 0); return 0; }</lang>

### Lexicographic ordered generation

Without recursions, generate all combinations in sequence. Basic logic: put n items in the first n of m slots; each step, if right most slot can be moved one slot further right, do so; otherwise find right most item that can be moved, move it one step and put all items already to its right next to it. <lang c>#include <stdio.h>

void comb(int m, int n, unsigned char *c) { int i; for (i = 0; i < n; i++) c[i] = n - i;

while (1) { for (i = n; i--;) printf("%d%c", c[i], i ? ' ': '\n');

/* this check is not strictly necessary, but if m is not close to n, it makes the whole thing quite a bit faster */ i = 0; if (c[i]++ < m) continue;

for (; c[i] >= m - i;) if (++i >= n) return; for (c[i]++; i; i--) c[i-1] = c[i] + 1; } }

int main() { unsigned char buf[100]; comb(5, 3, buf); return 0; }</lang>

## C#

<lang csharp>using System; using System.Collections.Generic;

public class Program {

   public static IEnumerable<int[]> Combinations(int m, int n)
{
int[] result = new int[m];
Stack<int> stack = new Stack<int>();
stack.Push(0);

           while (stack.Count > 0)
{
int index = stack.Count - 1;
int value = stack.Pop();

               while (value < n)
{
result[index++] = ++value;
stack.Push(value);

                   if (index == m)
{
yield return result;
break;
}
}
}
}

   static void Main()
{
foreach (int[] c in Combinations(3, 5))
{
Console.WriteLine(string.Join(",", c));
Console.WriteLine();
}
}


}</lang>

Here is another implementation that uses recursion, intead of an explicit stack:

<lang csharp> using System; using System.Collections.Generic;

public class Program {

 public static IEnumerable<int[]> FindCombosRec(int[] buffer, int done, int begin, int end)
{
for (int i = begin; i < end; i++)
{
buffer[done] = i;

     if (done == buffer.Length - 1)
yield return buffer;
else
foreach (int[] child in FindCombosRec(buffer, done+1, i+1, end))
yield return child;
}
}

 public static IEnumerable<int[]> FindCombinations(int m, int n)
{
return FindCombosRec(new int[m], 0, 0, n);
}

 static void Main()
{
foreach (int[] c in FindCombinations(3, 5))
{
for (int i = 0; i < c.Length; i++)
{
Console.Write(c[i] + " ");
}
Console.WriteLine();
}
}


} </lang>

Recursive version <lang csharp>using System; class Combinations {

 static int k = 3, n = 5;
static int [] buf = new int [k];

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

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


}</lang>

## C++

<lang cpp>#include <algorithm>

1. include <iostream>
2. include <string>

void comb(int N, int K) {

   std::string bitmask(K, 1); // K leading 1's
bitmask.resize(N, 0); // N-K trailing 0's

   // print integers and permute bitmask
do {
for (int i = 0; i < N; ++i) // [0..N-1] integers
{
if (bitmask[i]) std::cout << " " << i;
}
std::cout << std::endl;


}

int main() {

   comb(5, 3);


}</lang>

Output:
 0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4


## Clojure

<lang clojure>(defn combinations

 "If m=1, generate a nested list of numbers [0,n)
If m>1, for each x in [0,n), and for each list in the recursion on [x+1,n), cons the two"
[m n]
(letfn [(comb-aux


[m start] (if (= 1 m) (for [x (range start n)] (list x)) (for [x (range start n) xs (comb-aux (dec m) (inc x))] (cons x xs))))]

   (comb-aux m 0)))


(defn print-combinations

 [m n]
(doseq [line (combinations m n)]
(doseq [n line]
(printf "%s " n))
(printf "%n")))</lang>


The below code do not comply to the task described above. However, the combinations of n elements taken from m elements might be more natural to be expressed as a set of unordered sets of elements in Clojure using its Set data structure.

<lang clojure> (defn combinations

 "Generate the combinations of n elements from a list of [0..m)"
[m n]
(let [xs (range m)]
(loop [i (int 0) res #{#{}}]
(if (== i n)
res
(recur (+ 1 i)
(set (for [x xs r res
:when (not-any? #{x} r)]
(conj r x))))))))


</lang>

## CoffeeScript

Basic backtracking solution. <lang coffeescript> combinations = (n, p) ->

 return [ [] ] if p == 0
i = 0
combos = []
combo = []
while combo.length < p
if i < n
combo.push i
i += 1
else
break if combo.length == 0
i = combo.pop() + 1

if combo.length == p
combos.push clone combo
i = combo.pop() + 1
combos


clone = (arr) -> (n for n in arr)

N = 5 for i in [0..N]

 console.log "------ #{N} #{i}"
for combo in combinations N, i
console.log combo



</lang>

Output:
> coffee combo.coffee
------ 5 0
[]
------ 5 1
[ 0 ]
[ 1 ]
[ 2 ]
[ 3 ]
[ 4 ]
------ 5 2
[ 0, 1 ]
[ 0, 2 ]
[ 0, 3 ]
[ 0, 4 ]
[ 1, 2 ]
[ 1, 3 ]
[ 1, 4 ]
[ 2, 3 ]
[ 2, 4 ]
[ 3, 4 ]
------ 5 3
[ 0, 1, 2 ]
[ 0, 1, 3 ]
[ 0, 1, 4 ]
[ 0, 2, 3 ]
[ 0, 2, 4 ]
[ 0, 3, 4 ]
[ 1, 2, 3 ]
[ 1, 2, 4 ]
[ 1, 3, 4 ]
[ 2, 3, 4 ]
------ 5 4
[ 0, 1, 2, 3 ]
[ 0, 1, 2, 4 ]
[ 0, 1, 3, 4 ]
[ 0, 2, 3, 4 ]
[ 1, 2, 3, 4 ]
------ 5 5
[ 0, 1, 2, 3, 4 ]



## Common Lisp

<lang lisp>(defun map-combinations (m n fn)

 "Call fn with each m combination of the integers from 0 to n-1 as a list. The list may be destroyed after fn returns."
(let ((combination (make-list m)))
(labels ((up-from (low)
(let ((start (1- low)))
(lambda () (incf start))))
(mc (curr left needed comb-tail)
(cond
((zerop needed)
(funcall fn combination))
((= left needed)
(map-into comb-tail (up-from curr))
(funcall fn combination))
(t
(setf (first comb-tail) curr)
(mc (1+ curr) (1- left) (1- needed) (rest comb-tail))
(mc (1+ curr) (1- left) needed comb-tail)))))
(mc 0 n m combination))))</lang>

Output:
Example use
> (map-combinations 3 5 'print)

(0 1 2)
(0 1 3)
(0 1 4)
(0 2 3)
(0 2 4)
(0 3 4)
(1 2 3)
(1 2 4)
(1 3 4)
(2 3 4)
(2 3 4)

### Recursive method

<lang lisp>(defun comb (m list fn)

 (labels ((comb1 (l c m)


(when (>= (length l) m) (if (zerop m) (return-from comb1 (funcall fn c))) (comb1 (cdr l) c m) (comb1 (cdr l) (cons (first l) c) (1- m)))))

   (comb1 list nil m)))


(comb 3 '(0 1 2 3 4 5) #'print)</lang>

### Alternate, iterative method

<lang lisp>(defun next-combination (n a)

   (let ((k (length a)) m)
(loop for i from 1 do
(when (> i k) (return nil))
(when (< (aref a (- k i)) (- n i))
(setf m (aref a (- k i)))
(loop for j from i downto 1 do
(incf m)
(setf (aref a (- k j)) m))
(return t)))))


(defun all-combinations (n k)

   (if (or (< k 0) (< n k)) '()
(let ((a (make-array k)))
(loop for i below k do (setf (aref a i) i))
(loop collect (coerce a 'list) while (next-combination n a)))))



(defun map-combinations (n k fun)

   (if (and (>= k 0) (>= n k))
(let ((a (make-array k)))
(loop for i below k do (setf (aref a i) i))
(loop do (funcall fun (coerce a 'list)) while (next-combination n a)))))

all-combinations returns a list of lists

> (all-combinations 4 3) ((0 1 2) (0 1 3) (0 2 3) (1 2 3))

map-combinations applies a function to each combination

> (map-combinations 6 4 #'print) (0 1 2 3) (0 1 2 4) (0 1 2 5) (0 1 3 4) (0 1 3 5) (0 1 4 5) (0 2 3 4) (0 2 3 5) (0 2 4 5) (0 3 4 5) (1 2 3 4) (1 2 3 5) (1 2 4 5) (1 3 4 5) (2 3 4 5)</lang>

## Crystal

<lang Ruby> def comb(m, n)

   (0...n).to_a.each_combination(m) { |p| puts(p) }


end </lang>

<lang Bash> [0, 1, 2] [0, 1, 3] [0, 1, 4] [0, 2, 3] [0, 2, 4] [0, 3, 4] [1, 2, 3] [1, 2, 4] [1, 3, 4] [2, 3, 4] </lang>

## D

### Slow Recursive Version

Translation of: Python

<lang d>T[][] comb(T)(in T[] arr, in int k) pure nothrow {

   if (k == 0) return [[]];
typeof(return) result;
foreach (immutable i, immutable x; arr)
foreach (suffix; arr[i + 1 .. $].comb(k - 1)) result ~= x ~ suffix; return result;  } void main() {  import std.stdio; [0, 1, 2, 3].comb(2).writeln;  }</lang> Output: [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]] ### More Functional Recursive Version Translation of: Haskell Same output. <lang d>import std.stdio, std.algorithm, std.range; immutable(int)[][] comb(immutable int[] s, in int m) pure nothrow @safe {  if (!m) return [[]]; if (s.empty) return []; return s[1 ..$].comb(m - 1).map!(x => s[0] ~ x).array ~ s[1 .. $].comb(m);  } void main() {  4.iota.array.comb(2).writeln;  }</lang> ### Lazy Version <lang d>module combinations3; import std.traits: Unqual; struct Combinations(T, bool copy=true) {  Unqual!T[] pool, front; size_t r, n; bool empty = false; size_t[] indices; size_t len; bool lenComputed = false;   this(T[] pool_, in size_t r_) pure nothrow @safe { this.pool = pool_.dup; this.r = r_; this.n = pool.length; if (r > n) empty = true; indices.length = r; foreach (immutable i, ref ini; indices) ini = i; front.length = r; foreach (immutable i, immutable idx; indices) front[i] = pool[idx]; }   @property size_t length() /*logic_const*/ pure nothrow @nogc { static size_t binomial(size_t n, size_t k) pure nothrow @safe @nogc in { assert(n > 0, "binomial: n must be > 0."); } body { if (k < 0 || k > n) return 0; if (k > (n / 2)) k = n - k; size_t result = 1; foreach (size_t d; 1 .. k + 1) { result *= n; n--; result /= d; } return result; }   if (!lenComputed) { // Set cache. len = binomial(n, r); lenComputed = true; } return len; }   void popFront() pure nothrow @safe { if (!empty) { bool broken = false; size_t pos = 0; foreach_reverse (immutable i; 0 .. r) { pos = i; if (indices[i] != i + n - r) { broken = true; break; } } if (!broken) { empty = true; return; } indices[pos]++; foreach (immutable j; pos + 1 .. r) indices[j] = indices[j - 1] + 1; static if (copy) front = new Unqual!T[front.length]; foreach (immutable i, immutable idx; indices) front[i] = pool[idx]; } }  } Combinations!(T, copy) combinations(bool copy=true, T)  (T[] items, in size_t k)  in {  assert(items.length, "combinations: items can't be empty.");  } body {  return typeof(return)(items, k);  } // Compile with -version=combinations3_main to run main. version(combinations3_main) void main() {  import std.stdio, std.array, std.algorithm; [1, 2, 3, 4].combinations!false(2).array.writeln; [1, 2, 3, 4].combinations!true(2).array.writeln; [1, 2, 3, 4].combinations(2).map!(x => x).writeln;  }</lang> ### Lazy Lexicographical Combinations Includes an algorithm to find mth Lexicographical Element of a Combination. <lang d>module combinations4; import std.stdio, std.algorithm, std.conv; ulong choose(int n, int k) nothrow in {  assert(n >= 0 && k >= 0, "choose: no negative input.");  } body {  static ulong[][] cache;   if (n < k) return 0; else if (n == k) return 1; while (n >= cache.length) cache ~= [1UL]; // = choose(m, 0); auto kmax = min(k, n - k); while(kmax >= cache[n].length) { immutable h = cache[n].length; cache[n] ~= choose(n - 1, h - 1) + choose(n - 1, h); }   return cache[n][kmax];  } int largestV(in int p, in int q, in long r) nothrow in {  assert(p > 0 && q >= 0 && r >= 0, "largestV: no negative input.");  } body {  auto v = p - 1; while (choose(v, q) > r) v--; return v;  } struct Comb {  immutable int n, m;   @property size_t length() const /*nothrow*/ { return to!size_t(choose(n, m)); }   int[] opIndex(in size_t idx) const { if (m < 0 || n < 0) return []; if (idx >= length) throw new Exception("Out of bound"); ulong x = choose(n, m) - 1 - idx; int a = n, b = m; auto res = new int[m]; foreach (i; 0 .. m) { a = largestV(a, b, x); x = x - choose(a, b); b = b - 1; res[i] = n - 1 - a; } return res; }   int opApply(int delegate(ref int[]) dg) const { int[] yield;   foreach (i; 0 .. length) { yield = this[i]; if (dg(yield)) break; }   return 0; }   static auto On(T)(in T[] arr, in int m) { auto comb = Comb(arr.length, m);   return new class { @property size_t length() const /*nothrow*/ { return comb.length; }   int opApply(int delegate(ref T[]) dg) const { auto yield = new T[m];   foreach (c; comb) { foreach (idx; 0 .. m) yield[idx] = arr[c[idx]]; if (dg(yield)) break; }   return 0; } }; }  } version(combinations4_main)  void main() { foreach (c; Comb.On([1, 2, 3], 2)) writeln(c); }</lang>  ## E <lang e>def combinations(m, range) {  return if (m <=> 0) { [[]] } else { def combGenerator { to iterate(f) { for i in range { for suffix in combinations(m.previous(), range & (int > i)) { f(null, [i] + suffix) } } } } }  }</lang> ? for x in combinations(3, 0..4) { println(x) }  ## EchoLisp <lang scheme> using the native (combinations) function (lib 'list) (combinations (iota 5) 3) → ((0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4)) using an iterator (lib 'sequences) (take (combinator (iota 5) 3) #:all) → ((0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4)) defining a function (define (combine lst p) (cond  [(null? lst) null] [(< (length lst) p) null] [(= (length lst) p) (list lst)] [(= p 1) (map list lst)] [else (append (map cons (circular-list (first lst)) (combine (rest lst) (1- p))) (combine (rest lst) p))]))  (combine (iota 5) 3)  → ((0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4))  </lang> ## Egison <lang egison> (define$comb

 (lambda [$n$xs]
(match-all xs (list integer)
[(loop $i [1 ,n] <join _ <cons$a_i ...>> _) a])))


(test (comb 3 (between 0 4))) </lang>

Output:
{[|0 1 2|] [|0 1 3|] [|0 2 3|] [|1 2 3|] [|0 1 4|] [|0 2 4|] [|0 3 4|] [|1 2 4|] [|1 3 4|] [|2 3 4|]}


## Eiffel

The core of the program is the recursive feature solve, which returns all possible strings of length n with k "ones" and n-k "zeros". The strings are then evaluated, each resulting in k corresponding integers for the digits where ones are found. <lang Eiffel>

class COMBINATIONS

create make

feature

make (n, k: INTEGER) require n_positive: n > 0 k_positive: k > 0 k_smaller_equal: k <= n do create set.make set.extend ("") create sol.make sol := solve (set, k, n - k) sol := convert_solution (n, sol) ensure correct_num_of_sol: num_of_comb (n, k) = sol.count end

feature {None}

convert_solution (n: INTEGER; solution: LINKED_LIST [STRING]): LINKED_LIST [STRING] -- strings of 'k' digits between 1 and 'n' local i, j: INTEGER temp: STRING do create temp.make (n) from i := 1 until i > solution.count loop from j := 1 until j > n loop if solution [i].at (j) = '1' then temp.append (j.out) end j := j + 1 end solution [i].deep_copy (temp) temp.wipe_out i := i + 1 end Result := solution end

solve (seta: LINKED_LIST [STRING]; one, zero: INTEGER): LINKED_LIST [STRING] -- list of strings with a number of 'one' 1s and 'zero' 0, standig for wether the corresponing digit is taken or not. local new_P1, new_P0: LINKED_LIST [STRING] do create new_P1.make create new_P0.make if one > 0 then new_P1.deep_copy (seta) across new_P1 as P1 loop new_P1.item.append ("1") end new_P1 := solve (new_P1, one - 1, zero) end if zero > 0 then new_P0.deep_copy (seta) across new_P0 as P0 loop new_P0.item.append ("0") end new_P0 := solve (new_P0, one, zero - 1) end if one = 0 and zero = 0 then Result := seta else create Result.make Result.fill (new_p0) Result.fill (new_p1) end end

num_of_comb (n, k: INTEGER): INTEGER -- number of 'k' sized combinations out of 'n'. local upper, lower, m, l: INTEGER do upper := 1 lower := 1 m := n l := k from until m < n - k + 1 loop upper := m * upper lower := l * lower m := m - 1 l := l - 1 end Result := upper // lower end

end </lang> Test: <lang Eiffel> class APPLICATION

create make

feature

make do create comb.make (5, 3) across comb.sol as ar loop io.put_string (ar.item.out + "%T") end end

comb: COMBINATIONS

end

</lang>

Output:
345 245 235 234 145 135 134 125 124 123


## Elena

ELENA 3.4 : <lang elena>import system'routines. import extensions. import extensions'routines.

const int M = 3. const int N = 5.

numbers(n) [

   ^ Array new(n); populate(:i)<int>( i )


]

public program [

   var aNumbers := numbers(N).
Combinator new:M of:aNumbers; forEach(:aRow)
[
console printLine(aRow toLiteral)
].



]</lang>

Output:
0,1,2
0,1,3
0,1,4
0,2,3
0,2,4
0,3,4
1,2,3
1,2,4
1,3,4
2,3,4


## Elixir

Translation of: Erlang

<lang elixir>defmodule RC do

 def comb(0, _), do: [[]]
def comb(_, []), do: []
def comb(m, [h|t]) do
(for l <- comb(m-1, t), do: [h|l]) ++ comb(m, t)
end


end

{m, n} = {3, 5} list = for i <- 1..n, do: i Enum.each(RC.comb(m, list), fn x -> IO.inspect x end)</lang>

Output:
[1, 2, 3]
[1, 2, 4]
[1, 2, 5]
[1, 3, 4]
[1, 3, 5]
[1, 4, 5]
[2, 3, 4]
[2, 3, 5]
[2, 4, 5]
[3, 4, 5]


## Emacs Lisp

<lang elisp>(defun comb-recurse (m n n-max)

 (cond ((zerop m) '(()))
((= n-max n) '())
(t (append (mapcar #'(lambda (rest) (cons n rest))
(comb-recurse (1- m) (1+ n) n-max))
(comb-recurse m (1+ n) n-max)))))


(defun comb (m n)

 (comb-recurse m 0 n))


(comb 3 5)</lang>

Output:
((0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4))

## Erlang

<lang erlang> -module(comb). -compile(export_all).

comb(0,_) ->

   [[]];


comb(_,[]) ->

   [];


comb(N,[H|T]) ->

   [[H|L] || L <- comb(N-1,T)]++comb(N,T).


</lang>

### Dynamic Programming

Could be optimized with a custom zipwith/3 function instead of using lists:sublist/2.

<lang erlang> -module(comb). -export([combinations/2]).

combinations(K, List) ->

   lists:last(all_combinations(K, List)).


all_combinations(K, List) ->

   lists:foldr(
fun(X, Next) ->
Sub = lists:sublist(Next, length(Next) - 1),
Step = [[]] ++ [[[X|S] || S <- L] || L <- Sub],
lists:zipwith(fun lists:append/2, Step, Next)
end, [[[]]] ++ lists:duplicate(K, []), List).


</lang>

## ERRE

<lang ERRE> PROGRAM COMBINATIONS

CONST M_MAX=3,N_MAX=5

DIM COMBINATION[M_MAX],STACK[100,1]

PROCEDURE GENERATE(M)

  LOCAL I
IF (M>M_MAX) THEN
FOR I=1 TO M_MAX DO
PRINT(COMBINATION[I];" ";)
END FOR
PRINT
ELSE
FOR N=1 TO N_MAX DO
IF ((M=1) OR (N>COMBINATION[M-1])) THEN
COMBINATION[M]=N
! --- PUSH STACK -----------
STACK[SP,0]=M  STACK[SP,1]=N
SP=SP+1
! --------------------------

       GENERATE(M+1)

       ! --- POP STACK ------------
SP=SP-1
M=STACK[SP,0] N=STACK[SP,1]
! --------------------------
END IF
END FOR
END IF


END PROCEDURE

BEGIN

GENERATE(1)


END PROGRAM </lang>

Output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5


## F#

<lang fsharp>let choose m n =

   let rec fC prefix m from = seq {
let rec loopFor f = seq {
match f with
| [] -> ()
| x::xs ->
yield (x, fC [] (m-1) xs)
yield! loopFor xs
}
if m = 0 then yield prefix
else
for (i, s) in loopFor from do
for x in s do
yield prefix@[i]@x
}
fC [] m [0..(n-1)]


[<EntryPoint>] let main argv =

   choose 3 5
|> Seq.iter (printfn "%A")
0</lang>

Output:
[0; 1; 2]
[0; 1; 3]
[0; 1; 4]
[0; 2; 3]
[0; 2; 4]
[0; 3; 4]
[1; 2; 3]
[1; 2; 4]
[1; 3; 4]
[2; 3; 4]

## Factor

<lang factor>USING: math.combinatorics prettyprint ;

5 iota 3 all-combinations .</lang>

{
{ 0 1 2 }
{ 0 1 3 }
{ 0 1 4 }
{ 0 2 3 }
{ 0 2 4 }
{ 0 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 3 4 }
{ 2 3 4 }
}


This works with any kind of sequence: <lang factor>{ "a" "b" "c" } 2 all-combinations .</lang>

{ { "a" "b" } { "a" "c" } { "b" "c" } }

## Fortran

<lang fortran>program Combinations

 use iso_fortran_env
implicit none

 type comb_result
integer, dimension(:), allocatable :: combs
end type comb_result

 type(comb_result), dimension(:), pointer :: r
integer :: i, j

 call comb(5, 3, r)
do i = 0, choose(5, 3) - 1
do j = 2, 0, -1
write(*, "(I4, ' ')", advance="no") r(i)%combs(j)
end do
deallocate(r(i)%combs)
write(*,*) ""
end do
deallocate(r)


contains

 function choose(n, k, err)
integer :: choose
integer, intent(in) :: n, k
integer, optional, intent(out) :: err

   integer :: imax, i, imin, ie

   ie = 0
if ( (n < 0 ) .or. (k < 0 ) ) then
write(ERROR_UNIT, *) "negative in choose"
choose = 0
ie = 1
else
if ( n < k ) then
choose = 0
else if ( n == k ) then
choose = 1
else
imax = max(k, n-k)
imin = min(k, n-k)
choose = 1
do i = imax+1, n
choose = choose * i
end do
do i = 2, imin
choose = choose / i
end do
end if
end if
if ( present(err) ) err = ie
end function choose

 subroutine comb(n, k, co)
integer, intent(in) :: n, k
type(comb_result), dimension(:), pointer, intent(out) :: co

   integer :: i, j, s, ix, kx, hm, t
integer :: err

hm = choose(n, k, err)
if ( err /= 0 ) then
nullify(co)
return
end if

   allocate(co(0:hm-1))
do i = 0, hm-1
allocate(co(i)%combs(0:k-1))
end do
do i = 0, hm-1
ix = i; kx = k
do s = 0, n-1
if ( kx == 0 ) exit
t = choose(n-(s+1), kx-1)
if ( ix < t ) then
co(i)%combs(kx-1) = s
kx = kx - 1
else
ix = ix - t
end if
end do
end do

 end subroutine comb


end program Combinations</lang> Alternatively: <lang fortran>program combinations

 implicit none
integer, parameter :: m_max = 3
integer, parameter :: n_max = 5
integer, dimension (m_max) :: comb
character (*), parameter :: fmt = '(i0' // repeat (', 1x, i0', m_max - 1) // ')'

 call gen (1)


contains

 recursive subroutine gen (m)

   implicit none
integer, intent (in) :: m
integer :: n

   if (m > m_max) then
write (*, fmt) comb
else
do n = 1, n_max
if ((m == 1) .or. (n > comb (m - 1))) then
comb (m) = n
call gen (m + 1)
end if
end do
end if

 end subroutine gen


end program combinations</lang>

Output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5

## GAP

<lang gap># Built-in Combinations([1 .. n], m);

Combinations([1 .. 5], 3);

1. [ [ 1, 2, 3 ], [ 1, 2, 4 ], [ 1, 2, 5 ], [ 1, 3, 4 ], [ 1, 3, 5 ],
2. [ 1, 4, 5 ], [ 2, 3, 4 ], [ 2, 3, 5 ], [ 2, 4, 5 ], [ 3, 4, 5 ] ]</lang>

## Glee

<lang glee>5!3 >>> ,,\

$$(5!3) give all combinations of 3 out of 5$$(>>>) sorted up, (,,\) printed with crlf delimiters.</lang>

Result:

<lang glee>Result: 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5</lang>

## Go

<lang go>package main

import (

   "fmt"


)

func main() {

   comb(5, 3, func(c []int) {
fmt.Println(c)
})


}

func comb(n, m int, emit func([]int)) {

   s := make([]int, m)
last := m - 1
var rc func(int, int)
rc = func(i, next int) {
for j := next; j < n; j++ {
s[i] = j
if i == last {
emit(s)
} else {
rc(i+1, j+1)
}
}
return
}
rc(0, 0)


}</lang>

Output:
[0 1 2]
[0 1 3]
[0 1 4]
[0 2 3]
[0 2 4]
[0 3 4]
[1 2 3]
[1 2 4]
[1 3 4]
[2 3 4]

## Groovy

Following the spirit of the Haskell solution.

### In General

A recursive closure must be pre-declared. <lang groovy>def comb comb = { m, list ->

   def n = list.size()
m == 0 ?
[[]] :
(0..(n-m)).inject([]) { newlist, k ->
def sublist = (k+1 == n) ? [] : list[(k+1)..<n]
newlist += comb(m-1, sublist).collect { [list[k]] + it }
}


}</lang>

Test program: <lang groovy>def csny = [ "Crosby", "Stills", "Nash", "Young" ] println "Choose from ${csny}" (0..(csny.size())).each { i -> println "Choose${i}:"; comb(i, csny).each { println it }; println() }</lang>

Output:
Choose from [Crosby, Stills, Nash, Young]
Choose 0:
[]

Choose 1:
[Crosby]
[Stills]
[Nash]
[Young]

Choose 2:
[Crosby, Stills]
[Crosby, Nash]
[Crosby, Young]
[Stills, Nash]
[Stills, Young]
[Nash, Young]

Choose 3:
[Crosby, Stills, Nash]
[Crosby, Stills, Young]
[Crosby, Nash, Young]
[Stills, Nash, Young]

Choose 4:
[Crosby, Stills, Nash, Young]

### Zero-based Integers

<lang groovy>def comb0 = { m, n -> comb(m, (0..<n)) }</lang>

Test program: <lang groovy>println "Choose out of 5 (zero-based):" (0..3).each { i -> println "Choose ${i}:"; comb0(i, 5).each { println it }; println() }</lang> Output: Choose out of 5 (zero-based): Choose 0: [] Choose 1: [0] [1] [2] [3] [4] Choose 2: [0, 1] [0, 2] [0, 3] [0, 4] [1, 2] [1, 3] [1, 4] [2, 3] [2, 4] [3, 4] Choose 3: [0, 1, 2] [0, 1, 3] [0, 1, 4] [0, 2, 3] [0, 2, 4] [0, 3, 4] [1, 2, 3] [1, 2, 4] [1, 3, 4] [2, 3, 4] ### One-based Integers <lang groovy>def comb1 = { m, n -> comb(m, (1..n)) }</lang> Test program: <lang groovy>println "Choose out of 5 (one-based):" (0..3).each { i -> println "Choose${i}:"; comb1(i, 5).each { println it }; println() }</lang>

Output:
Choose out of 5 (one-based):
Choose 0:
[]

Choose 1:
[1]
[2]
[3]
[4]
[5]

Choose 2:
[1, 2]
[1, 3]
[1, 4]
[1, 5]
[2, 3]
[2, 4]
[2, 5]
[3, 4]
[3, 5]
[4, 5]

Choose 3:
[1, 2, 3]
[1, 2, 4]
[1, 2, 5]
[1, 3, 4]
[1, 3, 5]
[1, 4, 5]
[2, 3, 4]
[2, 3, 5]
[2, 4, 5]
[3, 4, 5]

It's more natural to extend the task to all (ordered) sublists of size m of a list.

Straightforward, unoptimized implementation with divide-and-conquer: <lang haskell>comb :: Int -> [a] -> a comb 0 _ = [[]] comb _ [] = [] comb m (x:xs) = map (x:) (comb (m-1) xs) ++ comb m xs</lang>

In the induction step, either x is not in the result and the recursion proceeds with the rest of the list xs, or it is in the result and then we only need m-1 elements.

Shorter version of the above: <lang haskell>import Data.List (tails)

comb :: Int -> [a] -> a comb 0 _ = [[]] comb m l = [x:ys | x:xs <- tails l, ys <- comb (m-1) xs]</lang>

To generate combinations of integers between 0 and n-1, use

<lang haskell>comb0 m n = comb m [0..n-1]</lang>

Similar, for integers between 1 and n, use

<lang haskell>comb1 m n = comb m [1..n]</lang>

Another method is to use the built in Data.List.subsequences function, filter for subsequences of length m and then sort:

<lang haskell>import Data.List (sort, subsequences) comb m n = sort . filter ((==m) . length) $subsequences [0..n-1]</lang> And yet another way is to use the list monad to generate all possible subsets: <lang haskell>comb m n = filter ((==m . length)$ filterM (const [True, False]) [0..n-1]</lang>

### Dynamic Programming

The first solution is inefficient because it repeatedly calculates the same subproblem in different branches of recursion. For example, comb m (x1:x2:xs) involves computing comb (m-1) (x2:xs) and comb m (x2:xs), both of which (separately) compute comb (m-1) xs. To avoid repeated computation, we can use dynamic programming:

<lang haskell>comb :: Int -> [a] -> a comb m xs = combsBySize xs !! m

where
combsBySize = foldr f ([[]] : repeat [])
f x next = zipWith (++) (map (map (x:)) ([]:next)) next</lang>


## Icon and Unicon

<lang Icon>procedure main() return combinations(3,5,0) end

procedure combinations(m,n,z) # demonstrate combinations /z := 1

write(m," combinations of ",n," integers starting from ",z) every put(L := [], z to n - 1 + z by 1) # generate list of n items from z write("Intial list\n",list2string(L)) write("Combinations:") every write(list2string(lcomb(L,m))) end

procedure list2string(L) # helper function every (s := "[") ||:= " " || (!L|"]") return s end

The provides the core procedure lcomb in lists written by Ralph E. Griswold and Richard L. Goerwitz.

<lang Icon>procedure lcomb(L,i) #: list combinations

  local j

  if i < 1 then fail
suspend if i = 1 then [!L]
else [L[j := 1 to *L - i + 1]] ||| lcomb(L[j + 1:0],i - 1)


end</lang>

Output:
3 combinations of 5 integers starting from 0
Intial list
[ 0 1 2 3 4 ]
Combinations:
[ 0 1 2 ]
[ 0 1 3 ]
[ 0 1 4 ]
[ 0 2 3 ]
[ 0 2 4 ]
[ 0 3 4 ]
[ 1 2 3 ]
[ 1 2 4 ]
[ 1 3 4 ]
[ 2 3 4 ]

## IS-BASIC

<lang IS-BASIC>100 PROGRAM "Combinat.bas" 110 LET MMAX=3:LET NMAX=5 120 NUMERIC COMB(0 TO MMAX) 130 CALL GENERATE(1) 140 DEF GENERATE(M) 150 NUMERIC N,I 160 IF M>MMAX THEN 170 FOR I=1 TO MMAX 180 PRINT COMB(I); 190 NEXT 200 PRINT 220 ELSE 230 FOR N=0 TO NMAX-1 240 IF M=1 OR N>COMB(M-1) THEN 250 LET COMB(M)=N 260 CALL GENERATE(M+1) 270 END IF 280 NEXT 290 END IF 300 END DEF</lang>

## J

### Library

<lang j>require'stats'</lang>

Example use:

<lang j> 3 comb 5 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4</lang>

All implementations here give that same result if given the same arguments.

### Iteration

 c=. 1 {.~ - d=. 1+y-x
z=. i.1 0
for_j. (d-1+y)+/&i.d do. z=. (c#j) ,. z{~;(-c){.&.><i.{.c=. +/\.c end.


)</lang>

### Recursion

 if. (x>:y)+.0=x do. i.(x<:y),x else. (0,.x combr&.<: y),1+x combr y-1 end.


)</lang> The M. uses memoization (caching) which greatly reduces the running time. As a result, this is probably the fastest of the implementations here.

### Brute Force

We can also generate all permutations and exclude those which are not properly sorted combinations. This is inefficient, but efficiency is not always important. <lang j>combb=: (#~ ((-:/:~)>/:~-:\:~)"1)@(# #: [: i. ^~)</lang>

## Java

Translation of: JavaScript
Works with: Java version 1.5+

public class Comb{

       public static void main(String[] args){
System.out.println(comb(3,5));
}

       public static String bitprint(int u){
String s= "";
for(int n= 0;u > 0;++n, u>>= 1)
if((u & 1) > 0) s+= n + " ";
return s;
}

       public static int bitcount(int u){
int n;
for(n= 0;u > 0;++n, u&= (u - 1));//Turn the last set bit to a 0
return n;
}

       public static LinkedList<String> comb(int c, int n){
for(int u= 0;u < 1 << n;u++)
if(bitcount(u) == c) s.push(bitprint(u));
Collections.sort(s);
return s;
}


}</lang>

## JavaScript

### Imperative

<lang javascript>function bitprint(u) {

 var s="";
for (var n=0; u; ++n, u>>=1)
if (u&1) s+=n+" ";
return s;


} function bitcount(u) {

 for (var n=0; u; ++n, u=u&(u-1));
return n;


} function comb(c,n) {

 var s=[];
for (var u=0; u<1<<n; u++)
if (bitcount(u)==c)
s.push(bitprint(u))
return s.sort();


} comb(3,5)</lang>

Alternative recursive version using and an array of values instead of length:

Translation of: Python

<lang javascript>function combinations(arr, k){

   var i,
subI,
ret = [],
sub,
next;
for(i = 0; i < arr.length; i++){
if(k === 1){
ret.push( [ arr[i] ] );
}else{
sub = combinations(arr.slice(i+1, arr.length), k-1);
for(subI = 0; subI < sub.length; subI++ ){
next = sub[subI];
next.unshift(arr[i]);
ret.push( next );
}
}
}
return ret;


} combinations([0,1,2,3,4], 3); // produces: [[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]

combinations(["Crosby", "Stills", "Nash", "Young"], 3); // produces: [["Crosby", "Stills", "Nash"], ["Crosby", "Stills", "Young"], ["Crosby", "Nash", "Young"], ["Stills", "Nash", "Young"]] </lang>

### Functional

#### ES5

Simple recursion:

<lang JavaScript>(function () {

 function comb(n, lst) {
if (!n) return [[]];
if (!lst.length) return [];

   var x = lst[0],
xs = lst.slice(1);

   return comb(n - 1, xs).map(function (t) {
return [x].concat(t);
}).concat(comb(n, xs));
}


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

 return comb(3, range(0, 4))

.map(function (x) {
return x.join(' ');
}).join('\n');


})();</lang>

We can significantly improve on the performance of the simple recursive function by deriving a memoized version of it, which stores intermediate results for repeated use.

<lang JavaScript>(function (n) {

 // n -> [a] -> a
function comb(n, lst) {
if (!n) return [[]];
if (!lst.length) return [];

   var x = lst[0],
xs = lst.slice(1);

   return comb(n - 1, xs).map(function (t) {
return [x].concat(t);
}).concat(comb(n, xs));
}

 // 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;
});
}

 var fnMemoized = memoized(comb),
lstRange = range(0, 4);

 return fnMemoized(n, lstRange)

 .map(function (x) {
return x.join(' ');
}).join('\n');


})(3);</lang>

Output:

<lang JavaScript>0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4</lang>

#### ES6

Memoizing: <lang JavaScript>(() => {

   // combinations :: Int -> [a] -> a
const combinations = (n, xs) => {
const cmb_ = (n, xs) => {
if (n < 1) return [
[]
];
if (xs.length === 0) return [];
const h = xs[0],
tail = xs.slice(1);
return cmb_(n - 1, tail)
.map(cons(h))
.concat(cmb_(n, tail));
};
return memoized(cmb_)(n, xs);
}

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

   // 2 or more arguments
// curry :: Function -> Function
const curry = (f, ...args) => {
const go = xs => xs.length >= f.length ? (f.apply(null, xs)) :
function () {
return go(xs.concat(Array.from(arguments)));
};
return go([].slice.call(args, 1));
};

   // cons :: a -> [a] -> [a]
const cons = curry((x, xs) => [x].concat(xs));

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

   // Derive a memoized version of a function
// memoized :: Function -> Function
const memoized = f => {
let m = {};
return function (x) {
let args = [].slice.call(arguments),
strKey = args.join('-'),
v = m[strKey];
return (
(v === undefined) &&
(m[strKey] = v = f.apply(null, args)),
v
);
}
};

   // show :: a -> String
const show = (...x) =>
JSON.stringify.apply(
null, x.length > 1 ? [x[0], null, x[1]] : x
);

   return show(
memoized(combinations)(3, enumFromTo(0, 4))
);


})();</lang>

Output:
[[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4],
[0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]

Or, more generically: <lang JavaScript>(() => {

   'use strict';

   // COMBINATIONS -----------------------------------------------------------

   // comb :: Int -> Int -> Int
const comb = (m, n) => combinations(m, enumFromTo(0, n - 1));

   // combinations :: Int -> [a] -> a
const combinations = (k, xs) =>
sort(filter(xs => k === xs.length, subsequences(xs)));


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

   // cons :: a -> [a] -> [a]
const cons = (x, xs) => [x].concat(xs);

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

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

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

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

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

   // sort :: Ord a => [a] -> [a]
const sort = xs => xs.sort();

   // stringChars :: String -> [Char]
const stringChars = s => s.split();

   // subsequences :: [a] -> a
const subsequences = xs => {

       // nonEmptySubsequences :: [a] -> a
const nonEmptySubsequences = xxs => {
if (isNull(xxs)) return [];
const [x, xs] = uncons(xxs);
const f = (r, ys) => cons(ys, cons(cons(x, ys), r));

           return cons([x], foldr(f, [], nonEmptySubsequences(xs)));
};

       return nonEmptySubsequences(
(typeof xs === 'string' ? stringChars(xs) : xs)
);
};

   // uncons :: [a] -> Maybe (a, [a])
const uncons = xs => xs.length ? [xs[0], xs.slice(1)] : undefined;


   // TEST -------------------------------------------------------------------
return show(
comb(3, 5)
);


})();</lang>

Output:
[[0,1,2],[0,1,3],[0,1,4],[0,2,3],[0,2,4],[0,3,4],[1,2,3],[1,2,4],[1,3,4],[2,3,4]]

## jq

combination(r) generates a stream of combinations of the input array. The stream can be captured in an array as shown in the second example. <lang jq>def combination(r):

 if r > length or r < 0 then empty
elif r == length then .
else  ( [.[0]] + (.[1:]|combination(r-1))),
( .[1:]|combination(r))
end;

1. select r integers from the set (0 .. n-1)

def combinations(n;r): [range(0;n)] | combination(r);</lang> Example 1

combinations(5;3)

Output:
[0,1,2]
[0,1,3]
[0,1,4]
[0,2,3]
[0,2,4]
[0,3,4]
[1,2,3]
[1,2,4]
[1,3,4]
[2,3,4]


Example 2

["a", "b", "c", "d", "e"] | combination(3) ] | length

Output:
10


## Julia

The combinations function in the Combinatorics.jl package generates an iterable sequence of the combinations that you can loop over. (Note that the combinations are computed on the fly during the loop iteration, and are not pre-computed or stored since there many be a very large number of them.) <lang julia>using Combinatorics n = 4 m = 3 for i in combinations(0:n,m)

   println(i')


end</lang>

Output:
[0 1 2]
[0 1 3]
[0 1 4]
[0 2 3]
[0 2 4]
[0 3 4]
[1 2 3]
[1 2 4]
[1 3 4]
[2 3 4]


Recursive solution without the library

The previous solution is the best: it is most elegant, production stile solution.

If, on the other hand we wanted to show how it could be done in Julia, this recursive solution shows some potentials of Julia lang. <lang julia>##############################

1. COMBINATIONS OF 3 OUT OF 5 #
1. Set n and m

m = 5 n = 3

1. Prepare the boundary of the calculation. Only m - n numbers are changing in each position.

max_n = m - n

1. Prepare an array for result

result = zeros(Int64, n)

function combinations(pos, val) # n, max_n and result are visible in the function

   for i = val:max_n                      # from current value to the boundary
result[pos] = pos + i              # fill the position of result
if pos < n                         # if combination isn't complete,
combinations(pos+1, i)         # go to the next position
else
println(result)                # combination is complete, print it
end
end


end

combinations(1, 0) end</lang>

Output:
[1, 2, 3]
[1, 2, 4]
[1, 2, 5]
[1, 3, 4]
[1, 3, 5]
[1, 4, 5]
[2, 3, 4]
[2, 3, 5]
[2, 4, 5]
[3, 4, 5]


## K

Recursive implementation:

<lang k>comb:{[n;k]

   f:{:[k=#x; :,x; :,/_f' x,'(1+*|x) _ !n]}
:,/f' !n


}</lang>

## Kotlin

Translation of: Pascal

<lang scala>class Combinations(val m: Int, val n: Int) {

   private val combination = IntArray(m)

   init {
generate(0)
}

   private fun generate(k: Int) {
if (k >= m) {
for (i in 0 until m) print("${combination[i]} ") println() } else { for (j in 0 until n) if (k == 0 || j > combination[k - 1]) { combination[k] = j generate(k + 1) } } }  } fun main(args: Array<String>) {  Combinations(3, 5)  }</lang> Output: 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4  ## Logo <lang logo>to comb :n :list  if :n = 0 [output [[]]] if empty? :list [output []] output sentence map [sentence first :list ?] comb :n-1 bf :list ~ comb :n bf :list  end print comb 3 [0 1 2 3 4]</lang> ## Lua <lang lua> function map(f, a, ...) if a then return f(a), map(f, ...) end end function incr(k) return function(a) return k > a and a or a+1 end end function combs(m, n)  if m * n == 0 then return {{}} end local ret, old = {}, combs(m-1, n-1) for i = 1, n do for k, v in ipairs(old) do ret[#ret+1] = {i, map(incr(i), unpack(v))} end end return ret  end for k, v in ipairs(combs(3, 5)) do print(unpack(v)) end </lang> ## M2000 Interpreter Including a helper sub to export result to clipboard through a global variable (a temporary global variable) <lang M2000 Interpreter> Module Checkit {  Global a$
Document a$Module Combinations (m as long, n as long){ Module Level (n, s, h) { If n=1 then { while Len(s) { Print h, car(s) ToClipBoard() s=cdr(s) } } Else { While len(s) { call Level n-1, cdr(s), cons(h, car(s)) s=cdr(s) } } Sub ToClipBoard() local m=each(h) Local b$=""
While m {
b$+=If$(Len(b$)<>0->" ","")+Format$("{0::-10}",Array(m))
}
b$+=If$(Len(b$)<>0->" ","")+Format$("{0::-10}",Array(s,0))+{
}
a$<=b$   ' assign to global need <=
End Sub
}
If m<1 or n<1 then Error
s=(,)
for i=0 to n-1 {
s=cons(s, (i,))
}
}
Clear a$Combinations 3, 5 ClipBoard a$


} Checkit </lang>

Output:
         0          1          2
0          1          3
0          1          4
0          2          3
0          2          4
0          3          4
1          2          3
1          2          4
1          3          4
2          3          4


### Step by Step

<lang M2000 Interpreter> Module StepByStep {

     Function CombinationsStep (a, nn) {
c1=lambda (&f, &a) ->{
=car(a) : a=cdr(a) : f=len(a)=0
}
m=len(a)
c=c1
n=m-nn+1
p=2
while m>n {
c1=lambda c2=c,n=p, z=(,) (&f, &m) ->{
if len(z)=0 then z=cdr(m)
=cons(car(m),c2(&f, &z))
if f then z=(,) : m=cdr(m) : f=len(m)+len(z)<n
}
c=c1
p++
m--
}
=lambda c, a (&f) -> {
=c(&f, &a)
}
}
k=false
StepA=CombinationsStep((1, 2, 3, 4,5), 3)
while not k {
Print StepA(&k)
}
k=false
StepA=CombinationsStep((0, 1, 2, 3, 4), 3)
while not k {
Print StepA(&k)
}
k=false
StepA=CombinationsStep(("A", "B", "C", "D","E"), 3)
while not k {
Print StepA(&k)
}
k=false
StepA=CombinationsStep(("CAT", "DOG", "BAT"), 2)
while not k {
Print StepA(&k)
}


} StepByStep </lang>

## M4

<lang M4>divert(-1) define(set',define($1[$2]',$3')') define(get',defn($1[$2]')') define(setrange',ifelse($3',',$2,define($1[$2],$3)'setrange($1,  incr($2),shift(shift(shift($@))))')')  define(for',  ifelse($#,0,$0, ifelse(eval($2<=$3),1, pushdef($1',$2)$4'popdef($1')$0($1',incr($2),$3,$4')')')')


define(show',

  for(k',0,decr($1),get(a,k) ')')  define(chklim',  ifelse(get(a',$3),eval($2-($1-$3)), chklim($1,$2,decr($3))',
set(a',$3,incr(get(a',$3)))'for(k',incr($3),decr($2),
set(a',k,incr(get(a',decr(k))))')'nextcomb($1,$2)')')


define(nextcomb',

  show($1)  ifelse(eval(get(a',0)<$2-$1),1,  chklim($1,$2,decr($1))')')


define(comb',

  for(j',0,decr($1),set(a',j,j)')'nextcomb($1,$2)')  divert comb(3,5)</lang> ## Maple This is built-in in Maple: <lang Maple>> combinat:-choose( 5, 3 ); [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5],  [2, 4, 5], [3, 4, 5]]  </lang> ## Mathematica <lang Mathematica>combinations[n_Integer, m_Integer]/;m>= 0:=Union[Sort /@ Permutations[Range[0, n - 1], {m}]]</lang> ## MATLAB This a built-in function in MATLAB called "nchoosek(n,k)". The argument "n" is a vector of values from which the combinations are made, and "k" is a scalar representing the amount of values to include in each combination. Task Solution: <lang MATLAB>>> nchoosek((0:4),3) ans =  0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4</lang>  ## Maxima <lang maxima>next_comb(n, p, a) := block(  [a: copylist(a), i: p], if a[1] + p = n + 1 then return(und), while a[i] - i >= n - p do i: i - 1, a[i]: a[i] + 1, for j from i + 1 thru p do a[j]: a[j - 1] + 1, a  )$

combinations(n, p) := block(

  [a: makelist(i, i, 1, p), v: [ ]],
while a # 'und do (v: endcons(a, v), a: next_comb(n, p, a)),
v


)$combinations(5, 3); /* [[1, 2, 3],  [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]] */</lang>  ## Modula-2 Translation of: Pascal Works with: ADW Modula-2 version any (Compile with the linker option Console Application). <lang modula2> MODULE Combinations; FROM STextIO IMPORT  WriteString, WriteLn;  FROM SWholeIO IMPORT  WriteInt;  CONST  MMax = 3; NMax = 5;  VAR  Combination: ARRAY [0 .. MMax] OF CARDINAL;  PROCEDURE Generate(M: CARDINAL); VAR  N, I: CARDINAL;  BEGIN  IF (M > MMax) THEN FOR I := 1 TO MMax DO WriteInt(Combination[I], 1); WriteString(' '); END; WriteLn; ELSE FOR N := 1 TO NMax DO IF (M = 1) OR (N > Combination[M - 1]) THEN Combination[M] := N; Generate(M + 1); END END END  END Generate; BEGIN  Generate(1);  END Combinations. </lang> Output: 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5  ## Nim <lang nim>iterator comb(m, n): seq[int] =  var c = newSeq[int](n) for i in 0 .. <n: c[i] = i   block outer: while true: yield c   var i = n-1 inc c[i] if c[i] <= m - 1: continue   while c[i] >= m - n + i: dec i if i < 0: break outer inc c[i] while i < n-1: c[i+1] = c[i] + 1 inc i  for i in comb(5, 3): echo i</lang> Output: @[0, 1, 2] @[0, 1, 3] @[0, 1, 4] @[0, 2, 3] @[0, 2, 4] @[0, 3, 4] @[1, 2, 3] @[1, 2, 4] @[1, 3, 4] @[2, 3, 4] Using explicit stack (deque) adopted from C#: <lang nim>iterator Combinations(m: int, n: int): seq[int] =  var result = newSeq[int](m) var stack = initDeque[int]() stack.addLast 0 while len(stack) > 0: var index = len(stack) - 1 var value = stack.popLast() while value < n: value = value + 1 result[index] = value index = index + 1 stack.addLast value   if index == m: yield result break  </lang> ## Octave <lang octave>nchoosek([0:4], 3)</lang> ## Oz This can be implemented as a trivial application of finite set constraints: <lang oz>declare  fun {Comb M N} proc {CombScript Comb} %% Comb is a subset of [0..N-1] Comb = {FS.var.upperBound {List.number 0 N-1 1}} %% Comb has cardinality M {FS.card Comb M} %% enumerate all possibilities {FS.distribute naive [Comb]} end in %% Collect all solutions and convert to lists {Map {SearchAll CombScript} FS.reflect.upperBoundList} end  in  {Inspect {Comb 3 5}}</lang>  ## PARI/GP <lang parigp>c(n,k,r,d)={  if(d==k, for(i=2,k+1, print1(r[i]" ")); print , for(i=r[d+1]+1,n, r[d+2]=i; c(n,k,r,d+1)));  } c(5,3,vector(5,i,i-1),0) </lang> ## Pascal <lang pascal>Program Combinations; const m_max = 3; n_max = 5;  var combination: array [0..m_max] of integer;  procedure generate(m: integer); var n, i: integer; begin if (m > m_max) then begin for i := 1 to m_max do write (combination[i], ' '); writeln; end else for n := 1 to n_max do if ((m = 1) or (n > combination[m-1])) then begin combination[m] := n; generate(m + 1); end; end;  begin generate(1);  end.</lang> Output: 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5  ## Perl The ntheory module has a combinations iterator that runs in lexicographic order. Library: ntheory <lang perl>use ntheory qw/forcomb/; forcomb { print "@_\n" } 5,3</lang> Output: 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4  Algorithm::Combinatorics also does lexicographic order and can return the whole array or an iterator: <lang perl>use Algorithm::Combinatorics qw/combinations/; my @c = combinations( [0..4], 3 ); print "@$_\n" for @c;</lang>

<lang perl>use Algorithm::Combinatorics qw/combinations/; my $iter = combinations([0..4],3); while (my$c = $iter->next) {  print "@$c\n";


}</lang>

Math::Combinatorics is another option but results will not be in lexicographic order as specified by the task.

## Perl5i

Use a recursive solution, derived from the Perl6 (Haskell) solution

• If we run out of eligable characters, we've gone too far, and won't find a solution along this path.
• If we are looking for a single character, each character in @set is elegible, so return each as the single element of an array.
• We have not yet reached the last character, so there are two possibilities:
1. push the first element of the set onto the front of an N-1 length combination from the remainder of the set.
2. skip the current element, and generate an N-length combination from the remainder

The major Perl5i -isms are the implicit "autoboxing" of the intermediate resulting array into an array object, with the use of unshift() as a method, and the "func" keyword and signature. Note that Perl can construct ranges of numbers or of letters, so it is natural to identify the characters as 'a' .. 'e'. <lang Perl5i> use perl5i::2;

1. ----------------------------------------
2. generate combinations of length $n consisting of characters 3. from the sorted set @set, using each character once in a 4. combination, with sorted strings in sorted order. 5. Returns a list of array references, each containing one combination. func combine($n, @set) {

 return unless @set;
return map { [ $_ ] } @set if$n == 1;

 my ($head) = shift @set; my @result = combine($n-1, @set );
for my $subarray ( @result ) {$subarray->unshift( $head ); } return ( @result, combine($n, @set ) );


}

say @$_ for combine( 3, ('a'..'e') ); </lang> Output: abc abd abe acd ace ade bcd bce bde cde  ## Perl 6 Works with: rakudo version 2015.12 There actually is a builtin: <lang perl6>.say for combinations(5,3);</lang> Output: (0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 3) (1 2 4) (1 3 4) (2 3 4) Here is an iterative routine with the same output: <lang perl6>sub combinations(Int$n, Int $k) {  return ([],) unless$k;
return if $k >$n || $n <= 0; my @c = ^$k;
gather loop {
take [@c];
next if @c[$k-1]++ <$n-1;
my $i =$k-2;
$i-- while$i >= 0 && @c[$i] >=$n-($k-$i);
last if $i < 0; @c[$i]++;
while ++$i <$k { @c[$i] = @c[$i-1] + 1; }
}


} .say for combinations(5,3);</lang>

## Phix

It does not get much simpler or easier than this. See Sudoku for a practical application of this algorithm <lang Phix>procedure comb(integer pool, needed, done=0, sequence chosen={})

   if needed=0 then    -- got a full set
?chosen         -- (or use a routine_id, result arg, or whatever)
return
end if
if done+needed>pool then return end if -- cannot fulfil
-- get all combinations with and without the next item:
done += 1
comb(pool,needed-1,done,append(chosen,done))
comb(pool,needed,done,chosen)


end procedure

comb(5,3)</lang>

Output:
{1,2,3}
{1,2,4}
{1,2,5}
{1,3,4}
{1,3,5}
{1,4,5}
{2,3,4}
{2,3,5}
{2,4,5}
{3,4,5}


## PHP

### non-recursive

Full non-recursive algorithm generating all combinations without repetions. Taken from here: [1]

Much slower than normal algorithm.

<lang php>

<?php


$a=array(1,2,3,4,5);$k=3; $n=5;$c=array_splice($a,$k); $b=array_splice($a, 0, $k);$j=$k-1; print_r($b);

       while (1) {

      		$m=array_search($b[$j]+1,$c);
if ($m!==false) { $c[$m]-=1; $b[$j]=$b[$j]+1; print_r($b);
}
if ($b[$k-1]==$n) { $i=$k-1; while ($i >= 0) {

if ($i == 0 &&$b[$i] ==$n-$k+1) break 2; $m=array_search($b[$i]+1,$c);  if ($m!==false) { $c[$m]=$c[$m]-1; $b[$i]=$b[$i]+1;

$g=$i; while ($g !=$k-1) { array_unshift ($c,$b[$g+1]);$b[$g+1]=$b[$g]+1;$g++; } $c=array_diff($c,$b); print_r($b); break;

      			 }


$i--; } } } ?> </lang> Output: Array ( [0] => 1 [1] => 2 ) Array ( [0] => 1 [1] => 3 ) Array ( [0] => 2 [1] => 3 )  ### recursive <lang php><?php function combinations_set($set = [], $size = 0) {  if ($size == 0) {
return [[]];
}

   if ($set == []) { return []; }  $prefix = [array_shift($set)];  $result = [];

   foreach (combinations_set($set,$size-1) as $suffix) {$result[] = array_merge($prefix,$suffix);
}

   foreach (combinations_set($set,$size) as $next) {$result[] = $next; }   return$result;


}

function combination_integer($n,$m) {

   return combinations_set(range(0, $n-1),$m);


}

assert(combination_integer(5, 3) == [

   [0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 2, 3],
[0, 2, 4],
[0, 3, 4],
[1, 2, 3],
[1, 2, 4],
[1, 3, 4],
[2, 3, 4]


]);

echo "3 comb 5:\n"; foreach (combination_integer(5, 3) as $combination) {  echo implode(", ",$combination), "\n";


} </lang>

Outputs:

3 comb 5:
0, 1, 2
0, 1, 3
0, 1, 4
0, 2, 3
0, 2, 4
0, 3, 4
1, 2, 3
1, 2, 4
1, 3, 4
2, 3, 4


## PicoLisp

Translation of: Scheme

<lang PicoLisp>(de comb (M Lst)

  (cond
((=0 M) '(NIL))
((not Lst))
(T
(conc
(mapcar
'((Y) (cons (car Lst) Y))
(comb (dec M) (cdr Lst)) )
(comb M (cdr Lst)) ) ) ) )


(comb 3 (1 2 3 4 5))</lang>

## Pop11

Natural recursive solution: first we choose first number i and then we recursively generate all combinations of m - 1 numbers between i + 1 and n - 1. Main work is done in the internal 'do_combs' function, the outer 'comb' just sets up variable to accumulate results and reverses the final result.

The 'el_lst' parameter to 'do_combs' contains partial combination (list of numbers which were chosen in previous steps) in reverse order.

<lang pop11>define comb(n, m);

   lvars ress = [];
define do_combs(l, m, el_lst);
lvars i;
if m = 0 then
cons(rev(el_lst), ress) -> ress;
else
for i from l to n - m do
do_combs(i + 1, m - 1, cons(i, el_lst));
endfor;
endif;
enddefine;
do_combs(0, m, []);
rev(ress);


enddefine;

comb(5, 3) ==></lang>

## PowerShell

An example of how PowerShell itself can translate C# code: <lang PowerShell> $source = @'  using System; using System.Collections.Generic;   namespace Powershell { public class CSharp { public static IEnumerable<int[]> Combinations(int m, int n) { int[] result = new int[m]; Stack<int> stack = new Stack<int>(); stack.Push(0); while (stack.Count > 0) { int index = stack.Count - 1; int value = stack.Pop(); while (value < n) { result[index++] = value++; stack.Push(value); if (index == m) { yield return result; break; } } } } } }  '@ Add-Type -TypeDefinition$source -Language CSharp

[Powershell.CSharp]::Combinations(3,5) | Format-Wide {$_} -Column 3 -Force </lang> Output: 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4  ## OpenEdge/Progress Translation of: Julia <lang OpenEdge/Progress> define variable r as integer no-undo extent 3. define variable m as integer no-undo initial 5. define variable n as integer no-undo initial 3. define variable max_n as integer no-undo. max_n = m - n. function combinations returns logical (input pos as integer, input val as integer):  define variable i as integer no-undo. do i = val to max_n: r[pos] = pos + i.  if pos lt n then combinations(pos + 1, i). else message r[1] - 1 r[2] - 1 r[3] - 1.  end.  end function. combinations(1, 0). </lang> Output: 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4 ## Prolog The solutions work with SWI-Prolog Solution with library clpfd : we first create a list of M elements, we say that the members of the list are numbers between 1 and N and there are in ascending order, finally we ask for a solution. <lang prolog>:- use_module(library(clpfd)). comb_clpfd(L, M, N) :-  length(L, M), L ins 1..N, chain(L, #<), label(L).</lang>  Output:  ?- comb_clpfd(L, 3, 5), writeln(L), fail. [1,2,3] [1,2,4] [1,2,5] [1,3,4] [1,3,5] [1,4,5] [2,3,4] [2,3,5] [2,4,5] [3,4,5] false. Another solution : <lang prolog>comb_Prolog(L, M, N) :-  length(L, M), fill(L, 1, N).  fill([], _, _). fill([H | T], Min, Max) :-  between(Min, Max, H), H1 is H + 1, fill(T, H1, Max).  </lang> with the same output. ### List comprehension Works with SWI-Prolog, library clpfd from Markus Triska, and list comprehension (see List comprehensions ). <lang Prolog>:- use_module(library(clpfd)). comb_lstcomp(N, M, V) :- V <- {L & length(L, N), L ins 1..M & all_distinct(L), chain(L, #<), label(L)}. </lang> Output: 2?- comb_lstcomp(3, 5, V). V = [[1,2,3],[1,2,4],[1,2,5],[1,3,4],[1,3,5],[1,4,5],[2,3,4],[2,3,5],[2,4,5],[3,4,5]] ; false.  ## Pure <lang pure>comb m n = comb m (0..n-1) with  comb 0 _ = [[]]; comb _ [] = []; comb m (x:xs) = [x:xs | xs = comb (m-1) xs] + comb m xs;  end; comb 3 5;</lang> ## PureBasic <lang PureBasic>Procedure.s Combinations(amount, choose)  NewList comb.s() ; all possible combinations with {amount} Bits For a = 0 To 1 << amount count = 0 ; count set bits For x = 0 To amount If (1 << x)&a count + 1 EndIf Next ; if set bits are equal to combination length ; we generate a String representing our combination and add it to list If count = choose string$ = ""
For x = 0 To amount
If (a >> x)&1
; replace x by x+1 to start counting with 1
String$+ Str(x) + " " EndIf Next AddElement(comb()) comb() = string$
EndIf
Next
; now we sort our list and format it for output as string
SortList(comb(), #PB_Sort_Ascending)
ForEach comb()
out$+ ", [ " + comb() + "]" Next ProcedureReturn Mid(out$, 3)


EndProcedure

Debug Combinations(5, 3)</lang>

## Pyret

<lang pyret>

fun combos<a>(lst :: List<a>, size :: Number) -> List<List<a>>:

 # return all subsets of lst of a certain size,
# maintaining the original ordering of the list

 # Let's handle a bunch of degenerate cases up front
# to be defensive...
if lst.length() < size:
# return an empty list if size is too big
[list:]
else if lst.length() == size:
# combos([list: 1,2,3,4]) == list[list: 1,2,3,4]]
[list: lst]
else if size == 1:
# combos(list: 5, 9]) == list[[list: 5], [list: 9]]
lst.map(lam(elem): [list: elem] end)
else:
# The main resursive step here is to consider
# all the combinations of the list that have the
# first element (aka head) and then those that don't
# don't.
cases(List) lst:
| empty => [list:]
# All the subsets of our list either include the
# first element of the list (aka head) or they don't.
with-head-combos = combos(rest, size - 1).map(
lam(combo):
)
end
end


where:

 # define semantics for the degenerate cases, although
# maybe we should just make some of these raise errors
combos([list:], 0) is [list: [list:]]
combos([list:], 1) is [list:]
combos([list: "foo"], 1) is [list: [list: "foo"]]
combos([list: "foo"], 2) is [list:]

 # test the normal stuff
lst = [list: 1, 2, 3]
combos(lst, 1) is [list:
[list: 1],
[list: 2],
[list: 3]
]
combos(lst, 2) is [list:
[list: 1, 2],
[list: 1, 3],
[list: 2, 3]
]
combos(lst, 3) is [list:
[list: 1, 2, 3]
]

 # remember the 10th row of Pascal's Triangle? :)
lst10 = [list: 1,2,3,4,5,6,7,8,9,10]
combos(lst10, 3).length() is 120
combos(lst10, 4).length() is 210
combos(lst10, 5).length() is 252
combos(lst10, 6).length() is 210
combos(lst10, 7).length() is 120

 # more sanity checks...
for each(sublst from combos(lst10, 6)):
sublst.length() is 6
end

 for each(sublst from combos(lst10, 9)):
sublst.length() is 9
end


end

fun int-combos(n :: Number, m :: Number) -> List<List<Number>>:

 doc: "return all lists of size m containing distinct, ordered nonnegative ints < n"
lst = range(0, n)
combos(lst, m)


where:

 int-combos(5, 5) is [list: [list: 0,1,2,3,4]]
int-combos(3, 2) is [list:
[list: 0, 1],
[list: 0, 2],
[list: 1, 2]
]


end

fun display-3-comb-5-for-rosetta-code():

 # The very concrete nature of this function is driven
# by the web page from Rosetta Code.  We want to display
#
# https://rosettacode.org/wiki/Combinations
results = int-combos(5, 3)
for each(lst from results):
print(lst.join-str(" "))
end


end

display-3-comb-5-for-rosetta-code()

</lang>

## Python

Starting from Python 2.6 and 3.0 you have a pre-defined function that returns an iterator. Here we turn the result into a list for easy printing: <lang python>>>> from itertools import combinations >>> list(combinations(range(5),3)) [(0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 3), (0, 2, 4), (0, 3, 4), (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]</lang>

Earlier versions could use functions like the following:

Translation of: E

<lang python>def comb(m, lst):

   if m == 0: return [[]]
return [[x] + suffix for i, x in enumerate(lst)
for suffix in comb(m - 1, lst[i + 1:])]</lang>


Example: <lang python>>>> comb(3, range(5)) [[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]</lang>

<lang python>def comb(m, s):

   if m == 0: return [[]]
if s == []: return []
return [s[:1] + a for a in comb(m-1, s[1:])] + comb(m, s[1:])


print comb(3, range(5))</lang>

## R

<lang R>print(combn(0:4, 3))</lang> Combinations are organized per column, so to provide an output similar to the one in the task text, we need the following: <lang R>r <- combn(0:4, 3) for(i in 1:choose(5,3)) print(r[,i])</lang>

## Racket

<lang racket> (define sublists

 (match-lambda**
[(0 _)           '(())]
[(_ '())         '()]
[(m (cons x xs)) (append (map (curry cons x) (sublists (- m 1) xs))
(sublists m xs))]))


(define (combinations n m)

 (sublists n (range m)))


</lang>

Output:
> (combinations 3 5)
'((0 1 2)
(0 1 3)
(0 1 4)
(0 2 3)
(0 2 4)
(0 3 4)
(1 2 3)
(1 2 4)
(1 3 4)
(2 3 4))


## REXX

This REXX program supports up to   61   symbols   (one symbol for each "thing").

It supports any number of "things" beyond the 61 symbols by using the actual number instead of a symbol. <lang rexx>/*REXX program displays combination sets for X things taken Y at a time. */ parse arg x y $. /*get optional arguments from the C.L. */ if x== | x=="," then x=5 /*No X specified? Then use default.*/ if y== | y=="," then y=3 /* " Y " " " " */ if$== | $=="," then$= '123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'

                                                /* [↑]  No  $specified? Use default.*/  say "────────────" x ' things taken ' y " at a time:" say "────────────" combN(x,y) ' combinations.' exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ combN: procedure expose$; parse arg x,y; xp=x+1; xm=xp-y;  !.=0

                               do i=1  for y; !.i=i;                            end /*i*/
do j=1;  L=;  do d=1  for y; L=L word(substr($,!.d,1) !.d, 1); end /*d*/ say L; !.y=!.y+1 if !.y==xp then if .combN(y-1) then leave end /*j*/ return j  .combN: procedure expose !. y xm; parse arg d; if d==0 then return 1; p=!.d  do u=d to y; !.u=p+1; if !.u==xm+u then return .combN(u-1); p=!.u end /*u*/ return 0</lang>  output when the following was specified: 5 3 01234 ──────────── 5 things taken 3 at a time: 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4 ──────────── 10 combinations.  output when the following was specified: 5 3 abcde ──────────── 5 things taken 3 at a time: a b c a b d a b e a c d a c e a d e b c d b c e b d e c d e ──────────── 10 combinations.  ## Ring <lang ring> 1. Project : Combinations n = 5 k = 3 temp = [] comb = [] num = com(n, k) while true  temp = [] for n = 1 to 3 tm = random(4) + 1 add(temp, tm) next bool1 = (temp[1] = temp[2]) and (temp[1] = temp[3]) and (temp[2] = temp[3]) bool2 = (temp[1] < temp[2]) and (temp[2] < temp[3]) if not bool1 and bool2 add(comb, temp) ok 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 comb = sortfirst(comb, 1) see showarray(comb) + nl func com(n, k)  res1 = 1 for n1 = n - k + 1 to n res1 = res1 * n1 next res2 = 1 for n2 = 1 to k res2 = res2 * n2 next res3 = res1/res2 return res3  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 2 3] [1 2 4] [1 2 5] [1 3 4] [1 3 5] [1 4 5] [2 3 4] [2 3 5] [2 4 5] [3 4 5]  ## Ruby Works with: Ruby version 1.8.7+ <lang ruby>def comb(m, n)  (0...n).to_a.combination(m).to_a  end comb(3, 5) # => [[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]</lang> ## Rust Works with: Rust version 0.9 <lang rust> fn comb<T: std::fmt::Default>(arr: &[T], n: uint) {  let mut incl_arr: ~[bool] = std::vec::from_elem(arr.len(), false); comb_intern(arr, n, incl_arr, 0);  } fn comb_intern<T: std::fmt::Default>(arr: &[T], n: uint, incl_arr: &mut [bool], index: uint) {  if (arr.len() < n + index) { return; } if (n == 0) { let mut it = arr.iter().zip(incl_arr.iter()).filter_map(|(val, incl)| if (*incl) { Some(val) } else { None } ); for val in it { print!("{} ", *val); } print("\n"); return; }   incl_arr[index] = true; comb_intern(arr, n-1, incl_arr, index+1); incl_arr[index] = false;   comb_intern(arr, n, incl_arr, index+1);  } fn main() {  let arr1 = ~[1, 2, 3, 4, 5]; comb(arr1, 3);   let arr2 = ~["A", "B", "C", "D", "E"]; comb(arr2, 3);  } </lang> ## Scala <lang scala>implicit def toComb(m: Int) = new AnyRef {  def comb(n: Int) = recurse(m, List.range(0, n)) private def recurse(m: Int, l: List[Int]): List[List[Int]] = (m, l) match { case (0, _) => List(Nil) case (_, Nil) => Nil case _ => (recurse(m - 1, l.tail) map (l.head :: _)) ::: recurse(m, l.tail) }  }</lang> Usage: scala> 3 comb 5 res170: List[List[Int]] = List(List(0, 1, 2), List(0, 1, 3), List(0, 1, 4), List(0, 2, 3), List(0, 2, 4), List(0, 3, 4), List(1, 2, 3), List(1, 2, 4), List(1, 3, 4), List(2, 3, 4))  Lazy version using iterators: <lang scala> def combs[A](n: Int, l: List[A]): Iterator[List[A]] = n match {  case _ if n < 0 || l.lengthCompare(n) < 0 => Iterator.empty case 0 => Iterator(List.empty) case n => l.tails.flatMap({ case Nil => Nil case x :: xs => combs(n - 1, xs).map(x :: _) }) }</lang>  Usage: scala> combs(3, (0 to 4).toList).toList res0: List[List[Int]] = List(List(0, 1, 2), List(0, 1, 3), List(0, 1, 4), List(0, 2, 3), List(0, 2, 4), List(0, 3, 4), List(1, 2, 3), List(1, 2, 4), List(1, 3, 4), List(2, 3, 4))  ### Dynamic programming Adapted from Haskell version: <lang scala> def combs[A](n: Int, xs: List[A]): Stream[List[A]] =  combsBySize(xs)(n)   def combsBySize[A](xs: List[A]): Stream[Stream[List[A]]] = { val z: Stream[Stream[List[A]]] = Stream(Stream(List())) ++ Stream.continually(Stream.empty) xs.toStream.foldRight(z)((a, b) => zipWith[Stream[List[A]]](_ ++ _, f(a, b), b)) }   def zipWith[A](f: (A, A) => A, as: Stream[A], bs: Stream[A]): Stream[A] = (as, bs) match { case (Stream.Empty, _) => Stream.Empty case (_, Stream.Empty) => Stream.Empty case (a #:: as, b #:: bs) => f(a, b) #:: zipWith(f, as, bs) }   def f[A](x: A, xsss: Stream[Stream[List[A]]]): Stream[Stream[List[A]]] = Stream.empty #:: xsss.map(_.map(x :: _))</lang>  Usage: combs(3, (0 to 4).toList).toList res0: List[List[Int]] = List(List(0, 1, 2), List(0, 1, 3), List(0, 1, 4), List(0, 2, 3), List(0, 2, 4), List(0, 3, 4), List(1, 2, 3), List(1, 2, 4), List(1, 3, 4), List(2, 3, 4))  ### Using Scala Standard Runtime Library #### Scala REPL <lang scala>scala>(0 to 4).combinations(3).toList res0: List[scala.collection.immutable.IndexedSeq[Int]] = List(Vector(0, 1, 2), Vector(0, 1, 3), Vector(0, 1, 4), Vector(0, 2, 3), Vector(0, 2, 4), Vector(0, 3, 4), Vector(1, 2, 3), Vector(1, 2, 4), Vector(1, 3, 4), Vector(2, 3, 4))</lang> #### Other environments Output: See it running in your browser by ScalaFiddle (JavaScript, non JVM) or by Scastie (JVM). ## Scheme Like the Haskell code: <lang scheme>(define (comb m lst)  (cond ((= m 0) '(())) ((null? lst) '()) (else (append (map (lambda (y) (cons (car lst) y)) (comb (- m 1) (cdr lst))) (comb m (cdr lst))))))  (comb 3 '(0 1 2 3 4))</lang> ## Seed7 <lang seed7>$ include "seed7_05.s7i";

const type: combinations is array array integer;

const func combinations: comb (in array integer: arr, in integer: k) is func

 result
var combinations: combResult is combinations.value;
local
var integer: x is 0;
var integer: i is 0;
var array integer: suffix is 0 times 0;
begin
if k = 0 then
combResult := 1 times 0 times 0;
else
for x key i range arr do
for suffix range comb(arr[succ(i) ..], pred(k)) do
combResult &:= [] (x) & suffix;
end for;
end for;
end if;
end func;


const proc: main is func

 local
var array integer: aCombination is 0 times 0;
var integer: element is 0;
begin
for aCombination range comb([] (0, 1, 2, 3, 4), 3) do
for element range aCombination do
end for;
writeln;
end for;
end func;</lang>

Output:
  0  1  2
0  1  3
0  1  4
0  2  3
0  2  4
0  3  4
1  2  3
1  2  4
1  3  4
2  3  4


## SETL

<lang SETL>print({0..4} npow 3);</lang>

## Sidef

### Built-in

<lang ruby>combinations(5, 3, {|*c| say c })</lang>

### Recursive

Translation of: Perl5i

<lang ruby>func combine(n, set) {

   set.len || return []
n == 1  && return set.map{[_]}

   var (head, result)
result = combine(n-1, [set...])

   for subarray in result {
}

   result + combine(n, set)


}

combine(3, @^5).each {|c| say c }</lang>

### Iterative

<lang ruby>func forcomb(callback, n, k) {

   if (k == 0) {
callback([])
return()
}

   if (k<0 || k>n || n==0) {
return()
}

   var c = @^k

   loop {
callback([c...])
c[k-1]++ < n-1 && next
var i = k-2
while (i>=0 && c[i]>=(n-(k-i))) {
--i
}
i < 0 && break
c[i]++
while (++i < k) {
c[i] = c[i-1]+1
}
}

   return()


}

forcomb({|c| say c }, 5, 3)</lang>

Output:
[0, 1, 2]
[0, 1, 3]
[0, 1, 4]
[0, 2, 3]
[0, 2, 4]
[0, 3, 4]
[1, 2, 3]
[1, 2, 4]
[1, 3, 4]
[2, 3, 4]


Works with: FriCAS
Works with: OpenAxiom
Works with: Axiom

[reverse subSet(5,3,i)$SGCF for i in 0..binomial(5,3)-1]  [[0,1,2], [0,1,3], [0,2,3], [1,2,3], [0,1,4], [0,2,4], [1,2,4], [0,3,4], [1,3,4], [2,3,4]] Type: List(List(Integer))  </lang> SGCF ==> SymmetricGroupCombinatoricFunctions ## Smalltalk Works with: Pharo Works with: Squeak <lang smalltalk> (0 to: 4) combinations: 3 atATimeDo: [ :x | Transcript cr; show: x printString]. "output on Transcript: 1. (0 1 2) 2. (0 1 3) 3. (0 1 4) 4. (0 2 3) 5. (0 2 4) 6. (0 3 4) 7. (1 2 3) 8. (1 2 4) 9. (1 3 4) 10. (2 3 4)" </lang> ## Standard ML <lang sml>fun comb (0, _ ) = [[]]  | comb (_, [] ) = [] | comb (m, x::xs) = map (fn y => x :: y) (comb (m-1, xs)) @ comb (m, xs)  comb (3, [0,1,2,3,4]);</lang> ## Stata <lang stata>program combin tempfile cp tempvar k gen k'=1 quietly save "cp'" rename 1' 1'1 forv i=2/2' { joinby k' using "cp'" rename 1' 1'i' quietly drop if 1'i'<=1'=i'-1' } sort 1'* end</lang> Example <lang stata>. set obs 5 . gen a=_n . combin a 3 . list  +--------------+ | a1 a2 a3 | |--------------| 1. | 1 2 3 | 2. | 1 2 4 | 3. | 1 2 5 | 4. | 1 3 4 | 5. | 1 3 5 | |--------------| 6. | 1 4 5 | 7. | 2 3 4 | 8. | 2 3 5 | 9. | 2 4 5 | 10. | 3 4 5 | +--------------+</lang>  ### Mata <lang stata>function combinations(n,k) { a = J(comb(n,k),k,.) u = 1..k for (i=1; 1; i++) { a[i,.] = u for (j=k; j>0; j--) { if (u[j]-j<n-k) break } if (j<1) return(a) u[j..k] = u[j]+1..u[j]+1+k-j } } combinations(5,3)</lang> Output  1 2 3 +-------------+ 1 | 1 2 3 | 2 | 1 2 4 | 3 | 1 2 5 | 4 | 1 3 4 | 5 | 1 3 5 | 6 | 1 4 5 | 7 | 2 3 4 | 8 | 2 3 5 | 9 | 2 4 5 | 10 | 3 4 5 | +-------------+ ## Swift <lang Swift>func addCombo(prevCombo: [Int], var pivotList: [Int]) -> [([Int], [Int])] {  return (0..<pivotList.count) .map { _ -> ([Int], [Int]) in (prevCombo + [pivotList.removeAtIndex(0)], pivotList) }  } func combosOfLength(n: Int, m: Int) -> Int {  return [Int](1...m) .reduce([([Int](), [Int](0..<n))]) { (accum, _) in accum.flatMap(addCombo) }.map {$0.0
}


}

println(combosOfLength(5, 3))</lang>

Output:
[[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]

## Tcl

ref[2] <lang tcl>proc comb {m n} {

   set set [list]
for {set i 0} {$i <$n} {incr i} {lappend set $i} return [combinations$set $m]  } proc combinations {list size} {  if {$size == 0} {
return [list [list]]
}
set retval {}
for {set i 0} {($i +$size) <= [llength $list]} {incr i} { set firstElement [lindex$list $i] set remainingElements [lrange$list [expr {$i + 1}] end] foreach subset [combinations$remainingElements [expr {$size - 1}]] { lappend retval [linsert$subset 0 $firstElement] } } return$retval


}

comb 3 5 ;# ==> {0 1 2} {0 1 3} {0 1 4} {0 2 3} {0 2 4} {0 3 4} {1 2 3} {1 2 4} {1 3 4} {2 3 4}</lang>

## TXR

TXR has repeating and non-repeating permutation and combination functions that produce lazy lists. They are generic over lists, strings and vectors. In addition, the combinations function also works over hashes.

Combinations and permutations are produced in lexicographic order (except in the case of hashes).

<lang txrlisp>(defun comb-n-m (n m)

 (comb (range* 0 n) m))


(put-line 3 comb 5 = @(comb-n-m 5 3))</lang>

Run:
\$ txr combinations.tl
3 comb 5 = ((0 1 2) (0 1 3) (0 1 4) (0 2 3) (0 2 4) (0 3 4) (1 2 4) (1 3 4) (2 3 4))

## Ursala

Most of the work is done by the standard library function choices, whose implementation is shown here for the sake of comparison with other solutions, <lang Ursala>choices = ^(iota@r,~&l); leql@a^& ~&al?\&! ~&arh2fabt2RDfalrtPXPRT</lang> where leql is the predicate that compares list lengths. The main body of the algorithm (~&arh2fabt2RDfalrtPXPRT) concatenates the results of two recursive calls, one of which finds all combinations of the required size from the tail of the list, and the other of which finds all combinations of one less size from the tail, and then inserts the head into each. choices generates combinations of an arbitrary set but not necessarily in sorted order, which can be done like this. <lang Ursala>#import std

1. import nat

combinations = @rlX choices^|(iota,~&); -< @p nleq+ ==-~rh</lang>

• The sort combinator (-<) takes a binary predicate to a function that sorts a list in order of that predicate.
• The predicate in this case begins by zipping its two arguments together with @p.
• The prefiltering operator -~ scans a list from the beginning until it finds the first item to falsify a predicate (in this case equality, ==) and returns a pair of lists with the scanned items satisfying the predicate on the left and the remaining items on the right.
• The rh suffix on the -~ operator causes it to return only the head of the right list as its result, which in this case will be the first pair of unequal items in the list.
• The nleq function then tests whether the left side of this pair is less than or equal to the right.
• The overall effect of using everything starting from the @p as the predicate to a sort combinator is therefore to sort a list of lists of natural numbers according to the order of the numbers in the first position where they differ.

test program: <lang Ursala>#cast %nLL

example = combinations(3,5)</lang>

Output:
<
<0,1,2>,
<0,1,3>,
<0,1,4>,
<0,2,3>,
<0,2,4>,
<0,3,4>,
<1,2,3>,
<1,2,4>,
<1,3,4>,
<2,3,4>>

## V

like scheme (using variables) <lang v>[comb [m lst] let

  [ [m zero?] [[[]]]
[lst null?] [[]]
[true] [m pred lst rest comb [lst first swap cons]  map
m lst rest comb concat]
] when].</lang>


Using destructuring view and stack not *pure at all <lang v>[comb

  [ [pop zero?] [pop pop [[]]]
[null?] [pop pop []]
[true] [ [m lst : [m pred lst rest comb [lst first swap cons]  map
m lst rest comb concat]] view i ]
] when].</lang>


Pure concatenative version <lang v>[comb

  [2dup [a b : a b a b] view].
[2pop pop pop].

  [ [pop zero?] [2pop [[]]]
[null?] [2pop []]
[true] [2dup [pred] dip uncons swapd comb [cons] map popd rollup rest comb concat]
] when].</lang>


Using it

|3 [0 1 2 3 4] comb
=[[0 1 2] [0 1 3] [0 1 4] [0 2 3] [0 2 4] [0 3 4] [1 2 3] [1 2 4] [1 3 4] [2 3 4]]


## VBA

<lang vb> Option Explicit Option Base 0 'Option Base 1

Private ArrResult

Sub test()

   'compute
Main_Combine 5, 3

'return
Dim j As Long, i As Long, temp As String
For i = LBound(ArrResult, 1) To UBound(ArrResult, 1)
temp = vbNullString
For j = LBound(ArrResult, 2) To UBound(ArrResult, 2)
temp = temp & " " & ArrResult(i, j)
Next
Debug.Print temp
Next
Erase ArrResult


End Sub

Private Sub Main_Combine(M As Long, N As Long) Dim MyArr, i As Long

   ReDim MyArr(M - 1)
If LBound(MyArr) > 0 Then ReDim MyArr(M) 'Case Option Base 1
For i = LBound(MyArr) To UBound(MyArr)
MyArr(i) = i
Next i
i = IIf(LBound(MyArr) > 0, N, N - 1)
ReDim ArrResult(i, LBound(MyArr))
Combine MyArr, N, LBound(MyArr), LBound(MyArr)
ReDim Preserve ArrResult(UBound(ArrResult, 1), UBound(ArrResult, 2) - 1)
'In VBA Excel we can use Application.Transpose instead of personal Function Transposition
ArrResult = Transposition(ArrResult)


End Sub

Private Sub Combine(MyArr As Variant, Nb As Long, Deb As Long, Ind As Long) Dim i As Long, j As Long, N As Long

   For i = Deb To UBound(MyArr, 1)
ArrResult(Ind, UBound(ArrResult, 2)) = MyArr(i)
N = IIf(LBound(ArrResult, 1) = 0, Nb - 1, Nb)
If Ind = N Then
ReDim Preserve ArrResult(UBound(ArrResult, 1), UBound(ArrResult, 2) + 1)
For j = LBound(ArrResult, 1) To UBound(ArrResult, 1)
ArrResult(j, UBound(ArrResult, 2)) = ArrResult(j, UBound(ArrResult, 2) - 1)
Next j
Else
Call Combine(MyArr, Nb, i + 1, Ind + 1)
End If
Next i


End Sub

Private Function Transposition(ByRef MyArr As Variant) As Variant Dim T, i As Long, j As Long

   ReDim T(LBound(MyArr, 2) To UBound(MyArr, 2), LBound(MyArr, 1) To UBound(MyArr, 1))
For i = LBound(MyArr, 1) To UBound(MyArr, 1)
For j = LBound(MyArr, 2) To UBound(MyArr, 2)
T(j, i) = MyArr(i, j)
Next j
Next i
Transposition = T
Erase T


End Function</lang>

Output:

If Option Base 0 :

 0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4

If Option Base 1 :

 1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5

## VBScript

<lang vb> Function Dec2Bin(n) q = n Dec2Bin = "" Do Until q = 0 Dec2Bin = CStr(q Mod 2) & Dec2Bin q = Int(q / 2) Loop Dec2Bin = Right("00000" & Dec2Bin,6) End Function

Sub Combination(n,k) Dim arr() ReDim arr(n-1) For h = 0 To n-1 arr(h) = h + 1 Next Set list = CreateObject("System.Collections.Arraylist") For i = 1 To 2^n bin = Dec2Bin(i) c = 0 tmp_combo = "" If Len(Replace(bin,"0","")) = k Then For j = Len(bin) To 1 Step -1 If CInt(Mid(bin,j,1)) = 1 Then tmp_combo = tmp_combo & arr(c) & "," End If c = c + 1 Next list.Add Mid(tmp_combo,1,(k*2)-1) End If Next list.Sort For l = 0 To list.Count-1 WScript.StdOut.Write list(l) WScript.StdOut.WriteLine Next End Sub

'Testing with n = 5 / k = 3 Call Combination(5,3) </lang>

Output:
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
3,4,5


## XPL0

<lang XPL0>code ChOut=8, CrLf=9, IntOut=11; def M=3, N=5; int A(N-1);

proc Combos(D, S); \Display all size M combinations of N in sorted order int D, S; \depth of recursion, starting value of N int I; [if D<M then \depth < size

     for I:= S to N-1 do
[A(D):= I;
Combos(D+1, I+1);
]


else [for I:= 0 to M-1 do

       [IntOut(0, A(I));  ChOut(0, ^ )];
CrLf(0);
];


];

Combos(0, 0)</lang>

Output:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4


## zkl

Translation of: OCaml

<lang zkl>fcn comb(k,seq){ // no repeats, seq is finite

  seq=seq.makeReadOnly();	// because I append to parts of seq
fcn(k,seq){
if(k<=0)    return(T(T));
if(not seq) return(T);
self.fcn(k-1,seq[1,*]).pump(List,seq[0,1].extend)


.extend(self.fcn(k,seq[1,*]));

  }(k,seq);


}</lang> <lang zkl>comb(3,"abcde".split("")).apply("concat")</lang>

Output:
L("abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde")