# 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.

*        Combinations              26/05/2016COMBINE  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=nLOOPI1   STC    R8,0(R7)           do i=1 to n; c(i)=n-i+1         LA     R7,1(R7)             @c(i)++         BCT    R8,LOOPI1          next iLOOPBIG  LA     R10,PG             big loop {------------------         LH     R1,N               n         LA     R7,C-1(R1)         @c(i)         LH     R6,N               i=nLOOPI2   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)           @[email protected]+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=1LOOPI3   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 iELOOPI3  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 iELOOPI4  B      LOOPBIG            big loop }------------------ELOOPBIG L      R13,4(0,R13)       epilog          LM     R14,R12,12(R13)    "         XR     R15,R15            "         BR     R14                exitM        DC     H'5'               <=input N        DC     H'3'               <=input C        DS     64X                array of 8 bit integersPG       DC     CL92' '            buffer                 YREGS         END    COMBINE
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


with Ada.Text_IO;  use Ada.Text_IO; 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;

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

type Five is range 0..4;

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
# -*- coding: utf-8 -*- # COMMENT REQUIRED BY "prelude_combinations_generative.a68"  MODE COMBDATA = ~;PROVIDES:# COMBDATA*=~* ## 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
File: test_combinations.a68
#!/usr/bin/a68g --script ## -*- coding: utf-8 -*- # CO REQUIRED BY "prelude_combinations.a68" CO  MODE COMBDATA = INT;#PROVIDES:## COMBDATA~=INT~ ## comb ~=int list ~#PR READ "prelude_combinations.a68" PR; 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);# FOR COMBDATALIST recombination data IN # comb gen combinations(m, test data list #) DO (#,##    (COMBDATALIST recombination)VOID:(    printf ((list fmt, recombination, $l$))# OD # )))
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

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 rend 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 trueend next_comb return comb(5, 3)
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}}

### Functional composition

Translation of: JavaScript
-- 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 ifend 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} & xsend cons -- curry :: (Script|Handler) -> Scripton curry(f)    script        on |λ|(a)            script                on |λ|(b)                    |λ|(a, b) of mReturn(f)                end |λ|            end script        end |λ|    end scriptend 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 lstend enumFromTo -- intercalate :: Text -> [Text] -> Texton 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 strJoinedend intercalate -- isNull :: [a] -> Boolon isNull(xs)    if class of xs is string then        xs = ""    else        xs = {}    end ifend 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 tellend map -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend 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 ifend uncons -- unwords :: [String] -> Stringon unwords(xs)    intercalate(space, xs)end unwords
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

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   }}

## 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}

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


      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$DEF FNfact(N%) IF N%<=1 THEN = 1 ELSE = N%*FNfact(N%-1)  ## Bracmat The program first constructs a pattern with m variables and an expression that evaluates m variables into a combination. Then the program constructs a list of the integers 0 ... n-1. The real work is done in the expression !list:!pat. When a combination is found, it is added to the list of combinations. Then we force the program to backtrack and find the next combination by evaluating the always failing ~. When all combinations are found, the pattern fails and we are in the rhs of the last | operator. (comb= bvar combination combinations list m n pat pvar var. !arg:(?m.?n) & ( pat = ? & !combinations (.!combination):?combinations & ~ ) & :?list:?combination:?combinations & whl ' ( !m+-1:~<0:?m & chu$(utf$a+!m):?var & glf$('(%@?.$var)):(=?pvar) & '(? ()$pvar ()$pat):(=?pat) & glf$('(!.$var)):(=?bvar) & ( '$combination:(=)            & '$bvar:(=?combination) | '($bvar ()$combination):(=?combination) ) ) & whl ' (!n+-1:~<0:?n&!n !list:?list) & !list:!pat | !combinations); comb$(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

#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;}

### 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.

#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;}

## C#

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();        }    }}

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

 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();    }  }} 

Recursive version

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));    }  }}

## C++

#include <algorithm>#include <iostream>#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;    } while (std::prev_permutation(bitmask.begin(), bitmask.end()));} int main(){    comb(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


## 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")))

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.

 (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)))))))) 

## CoffeeScript

Basic backtracking solution.

 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 = 5for i in [0..N]  console.log "------ #{N} #{i}"  for combo in combinations N, i    console.log combo  
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

(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))))
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

(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)

### Alternate, iterative method

(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)

## Crystal

 def comb(m, n)    (0...n).to_a.each_combination(m) { |p| puts(p) }end 
 [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] 

## D

### Slow Recursive Version

Translation of: Python
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 .. $].comb(k - 1)) result ~= x ~ suffix; return result;} void main() { import std.stdio; [0, 1, 2, 3].comb(2).writeln;} Output: [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]] ### More Functional Recursive Version Translation of: Haskell Same output. import std.stdio, std.algorithm, std.range; immutable(int)[][] comb(immutable int[] s, in int m) pure nothrow @safe { if (!m) return [[]]; if (s.empty) return []; return s[1 ..$].comb(m - 1).map!(x => s[0] ~ x).array ~ s[1 .. $].comb(m);} void main() { 4.iota.array.comb(2).writeln;} ### Lazy Version 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;} ### Lazy Lexicographical Combinations Includes an algorithm to find mth Lexicographical Element of a Combination. module combinations4;import std.stdio, std.algorithm, std.conv; ulong choose(int n, int k) nothrowin { 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) nothrowin { 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); } ## 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) } } } } }} ? for x in combinations(3, 0..4) { println(x) }  ## EchoLisp  ;;;; 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))  ## 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))) 
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.

  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 	sol: LINKED_LIST [STRING] feature {None} 	set: LINKED_LIST [STRING] 	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 

Test:

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


## Elena

ELENA 3.4 :

import system'routines.import extensions.import extensions'routines. const int M = 3.const int N = 5.  numbers = (:anN)[    ^ Array new(anN); populate(:n)<int>( n )]. public program =[    var aNumbers := numbers(N).        Combinator new:M of:aNumbers; forEach(:aRow)    [        console printLine(aRow toLiteral)    ].     console readChar.].
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
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)  endend {m, n} = {3, 5}list = for i <- 1..n, do: iEnum.each(RC.comb(m, list), fn x -> IO.inspect x end)
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

(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)
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

 -module(comb).-compile(export_all). comb(0,_) ->    [[]];comb(_,[]) ->    [];comb(N,[H|T]) ->    [[H|L] || L <- comb(N-1,T)]++comb(N,T). 

### Dynamic Programming

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

 -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). 

## 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 IFEND PROCEDURE BEGIN GENERATE(1)END PROGRAM 
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#

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 [email protected][i]@x            }    fC [] m [0..(n-1)] [<EntryPoint>]let main argv =     choose 3 5    |> Seq.iter (printfn "%A")    0
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

USING: math.combinatorics prettyprint ; 5 iota 3 all-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 }
}


This works with any kind of sequence:

{ "a" "b" "c" } 2 all-combinations .
{ { "a" "b" } { "a" "c" } { "b" "c" } }

## 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

Alternatively:

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
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

# Built-inCombinations([1 .. n], m); Combinations([1 .. 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 ] ]

## Glee

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

Result:

Result:1 2 31 2 41 2 51 3 41 3 51 4 52 3 42 3 52 4 53 4 5

## 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)}
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.

def combcomb = { 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 }        }}

Test program:

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() }
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

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

Test program:

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

comb :: Int -> [a] -> [[a]]comb 0 _      = [[]]comb _ []     = []comb m (x:xs) = map (x:) (comb (m-1) xs) ++ comb m xs

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:

import Data.List (tails) comb :: Int -> [a] -> [[a]]comb 0 _      = [[]]comb m l = [x:ys | x:xs <- tails l, ys <- comb (m-1) xs]

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

comb0 m n = comb m [0..n-1]

Similar, for integers between 1 and n, use

comb1 m n = comb m [1..n]

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

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

### 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:

comb :: Int -> [a] -> [[a]]comb m xs = combsBySize xs !! m where   combsBySize = foldr f ([[]] : repeat [])   f x next = zipWith (++) (map (map (x:)) ([]:next)) next

## Icon and Unicon

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 zwrite("Intial list\n",list2string(L))write("Combinations:")every write(list2string(lcomb(L,m)))end procedure list2string(L)                           # helper functionevery (s := "[") ||:= " " || (!L|"]")return send link lists
The provides the core procedure lcomb in lists written by Ralph E. Griswold and Richard L. Goerwitz.
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
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 ]

## J

### Library

require'stats'

Example use:

   3 comb 50 1 20 1 30 1 40 2 30 2 40 3 41 2 31 2 41 3 42 3 4

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

### Iteration

comb1=: dyad define  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.)

### Recursion

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

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.

combb=: (#~ ((-:/:~)>/:~-:\:~)"1)@(# #: [: i. ^~)

## Java

Translation of: JavaScript
Works with: Java version 1.5+
import java.util.Collections;import java.util.LinkedList; 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){                LinkedList<String> s= new LinkedList<String>();                for(int u= 0;u < 1 << n;u++)                        if(bitcount(u) == c) s.push(bitprint(u));                Collections.sort(s);                return s;        }}

## JavaScript

### Imperative

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)

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

Translation of: Python
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"]] 

### Functional

#### ES5

Simple recursion:

(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'); })();

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.

(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);

Output:
0 1 20 1 30 1 40 2 30 2 40 3 41 2 31 2 41 3 42 3 4

#### ES6

Memoizing:

(() => {     // 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))    );})();
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:

(() => {    '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)    );})();
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.

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; # select r integers from the set (0 .. n-1)def combinations(n;r): [range(0;n)] | combination(r);

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.)

using Combinatoricsn = 4m = 3for i in combinations(0:n,m)    println(i')end
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.

############################### COMBINATIONS OF 3 OUT OF 5 ############################### # Set n and mm = 5n = 3 # Prepare the boundary of the calculation. Only m - n numbers are changing in each position.    max_n = m - n #Prepare an array for resultresult = 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   endend combinations(1, 0)end
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:

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

## Kotlin

Translation of: Pascal
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)} 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 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 :listendprint comb 3 [0 1 2 3 4] ## Lua  function map(f, a, ...) if a then return f(a), map(f, ...) end endfunction incr(k) return function(a) return k > a and a or a+1 end endfunction 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 retend for k, v in ipairs(combs(3, 5)) do print(unpack(v)) end  ## 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($2),shift(shift(shift([email protected]))))')')define(for', ifelse($#,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,$2)')divert comb(3,5) ## Maple This is built-in in Maple: > combinat:-choose( 5, 3 );[[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]  ## Mathematica combinations[n_Integer, m_Integer]/;m>= 0:=Union[Sort /@ Permutations[Range[0, n - 1], {m}]] ## MATLAB This a built-in function in MATLAB called "nchoosek(n,k)". The argument "n" is a vector of values from which the combinations are made, and "k" is a scalar representing the amount of values to include in each combination. Task Solution: >> nchoosek((0:4),3) ans = 0 1 2 0 1 3 0 1 4 0 2 3 0 2 4 0 3 4 1 2 3 1 2 4 1 3 4 2 3 4 ## Maxima next_comb(n, p, a) := block( [a: copylist(a), i: p], if a[1] + p = n + 1 then return(und), while a[i] - i >= n - p do i: i - 1, a[i]: a[i] + 1, for j from i + 1 thru p do a[j]: a[j - 1] + 1, a)$ 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]] */ ## Modula-2 Translation of: Pascal Works with: ADW Modula-2 version any (Compile with the linker option Console Application).  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 ENDEND Generate; BEGIN Generate(1);END Combinations.  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 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 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] ## OCaml Like the Haskell code: let rec comb m lst = match m, lst with 0, _ -> [[]] | _, [] -> [] | m, x :: xs -> List.map (fun y -> x :: y) (comb (pred m) xs) @ comb m xs;;comb 3 [0;1;2;3;4];; Dynamic Programming solution: let comb m xs = let xs = Array.of_list xs in if m > Array.length xs then [] else begin let arr = Array.make (m+1) [] in arr.(0) <- [[]]; for j = 0 to Array.length xs - m do for i = 1 to m do arr.(i) <- arr.(i) @ List.map (fun ys -> xs.(j+i-1)::ys) arr.(i-1) done done; arr.(m) end;;comb 3 [0;1;2;3;4];; ## Octave nchoosek([0:4], 3) ## Oz This can be implemented as a trivial application of finite set constraints: 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} endin {Inspect {Comb 3 5}} ## PARI/GP 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)  ## 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. 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 use ntheory qw/forcomb/;forcomb { print "@_\n" } 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  Algorithm::Combinatorics also does lexicographic order and can return the whole array or an iterator: use Algorithm::Combinatorics qw/combinations/;my @c = combinations( [0..4], 3 );print "@$_\n" for @c;
use Algorithm::Combinatorics qw/combinations/;my $iter = combinations([0..4],3);while (my$c = $iter->next) { print "@$c\n";}

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'.

## Phix

It does not get much simpler or easier than this. See Sudoku for a practical application of this algorithm

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)
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.

  <?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;         			 }	 	$i--; } } } ?>  Output: Array ( [0] => 1 [1] => 2 ) Array ( [0] => 1 [1] => 3 ) Array ( [0] => 2 [1] => 3 )  ### recursive <?php function combinations_set($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";}

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
(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))

## 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.

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) ==>

## PowerShell

An example of how PowerShell itself can translate C# code:

## 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:]      | link(head, rest) =>        # 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):          link(head, combo) end          )        without-head-combos = combos(rest, size)        with-head-combos._plus(without-head-combos)    end  endwhere:  # 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  endend 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  # output similar to the top of this page:  #  # https://rosettacode.org/wiki/Combinations  results = int-combos(5, 3)  for each(lst from results):    print(lst.join-str(" "))  endend display-3-comb-5-for-rosetta-code()

## 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:

>>> 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)]

Earlier versions could use functions like the following:

Translation of: E
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:])]

Example:

>>> 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]]
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))

## R

print(combn(0:4, 3))

Combinations are organized per column, so to provide an output similar to the one in the task text, we need the following:

r <- combn(0:4, 3)for(i in 1:choose(5,3)) print(r[,i])

## 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)))
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.

/*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 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  # Project : Combinations n = 5k = 3temp = []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 okendcomb = 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  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+ def comb(m, n) (0...n).to_a.combination(m).to_aend 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]] ## Rust Works with: Rust version 0.9  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);}  ## 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) }} 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:  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 :: _) }) } 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:  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 :: _)) 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 scala>(0 to 4).combinations(3).toListres0: 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)) #### Other environments Output: See it running in your browser by ScalaFiddle (JavaScript, non JVM) or by Scastie (JVM). ## Scheme Like the Haskell code: (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)) ## 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        write(element lpad 3);      end for;      writeln;    end for;  end func;
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

print({0..4} npow 3);

## Sidef

### Built-in

combinations(5, 3, {|*c| say c })

### Recursive

Translation of: Perl5i
func combine(n, set) {     set.len || return []    n == 1  && return set.map{[_]}     var (head, result)    head   = set.shift    result = combine(n-1, [set...])     for subarray in result {        subarray.prepend(head)    }     result + combine(n, set)} combine(3, @^5).each {|c| say c }

### Iterative

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)
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))  SGCF ==> SymmetricGroupCombinatoricFunctions ## Smalltalk Works with: Pharo Works with: Squeak  (0 to: 4) combinations: 3 atATimeDo: [ :x | Transcript cr; show: x printString]. "output on Transcript:#(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)"  ## Standard ML 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]); ## Stata program combintempfile cptempvar kgen k'=1quietly save "cp'"rename 1' 1'1forv i=2/2' { joinby k' using "cp'" rename 1' 1'i' quietly drop if 1'i'<=1'=i'-1'}sort 1'*end Example . 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 | +--------------+ ### Mata 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) 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 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))
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]

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}

## 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).

(defun comb-n-m (n m)  (comb (range* 0 n) m)) (put-line 3 comb 5 = @(comb-n-m 5 3))
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,

choices = ^([email protected],~&l); [email protected]^& ~&al?\&! ~&arh2fabt2RDfalrtPXPRT

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.

#import std#import nat combinations = @rlX choices^|(iota,~&); -< @p nleq+ ==-~rh
• 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:

#cast %nLL example = combinations(3,5)
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)

[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].

Using destructuring view and stack not *pure at all

[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].

Pure concatenative version

[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].

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

 Option ExplicitOption Base 0'Option Base 1Private 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 ArrResultEnd 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 iEnd Sub Private Function Transposition(ByRef MyArr As Variant) As VariantDim 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 TEnd Function
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

 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	NextEnd Sub 'Testing with n = 5 / k = 3Call Combination(5,3)
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

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 orderint  D, S;              \depth of recursion, starting value of Nint  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)
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
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);}
comb(3,"abcde".split("")).apply("concat")
Output:
L("abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde")