Evaluate binomial coefficients: Difference between revisions

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Latest revision as of 12:34, 27 April 2024

This programming task, is to calculate ANY binomial coefficient.

However, it has to be able to output ${\binom {5}{3}}$ , which is 10.

This formula is recommended:

{\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}

See Also:

The number of samples of size k from n objects.

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

ABAP

CLASS lcl_binom DEFINITION CREATE PUBLIC.

  PUBLIC SECTION.
    CLASS-METHODS:
      calc
        IMPORTING n               TYPE i
                  k               TYPE i
        RETURNING VALUE(r_result) TYPE f.

ENDCLASS.

CLASS lcl_binom IMPLEMENTATION.

  METHOD calc.

    r_result = 1.
    DATA(i) = 1.
    DATA(m) = n.

    WHILE i <= k.
      r_result = r_result * m / i.
      i = i + 1.
      m = m - 1.
    ENDWHILE.

  ENDMETHOD.

ENDCLASS.

Output:

lcl_binom=>calc( n = 5 k = 3 )
1,0000000000000000E+01
lcl_binom=>calc( n = 60 k = 30 )
1,1826458156486142E+17

ACL2

(defun fac (n)
   (if (zp n)
       1
       (* n (fac (1- n)))))

(defun binom (n k)
   (/ (fac n) (* (fac (- n k)) (fac k)))

Ada

with Ada.Text_IO;  use Ada.Text_IO;
procedure Test_Binomial is
   function Binomial (N, K : Natural) return Natural is
      Result : Natural := 1;
      M      : Natural;
   begin
      if N < K then
         raise Constraint_Error;
      end if;
      if K > N/2 then -- Use symmetry
         M := N - K;
      else
         M := K;
      end if;
      for I in 1..M loop
         Result := Result * (N - M + I) / I;
      end loop;
      return Result;
   end Binomial;
begin
   for N in 0..17 loop
      for K in 0..N loop
         Put (Integer'Image (Binomial (N, K)));
      end loop;
      New_Line;
   end loop;
end Test_Binomial;

Output:

 1
 1 1
 1 2 1
 1 3 3 1
 1 4 6 4 1
 1 5 10 10 5 1
 1 6 15 20 15 6 1
 1 7 21 35 35 21 7 1
 1 8 28 56 70 56 28 8 1
 1 9 36 84 126 126 84 36 9 1
 1 10 45 120 210 252 210 120 45 10 1
 1 11 55 165 330 462 462 330 165 55 11 1
 1 12 66 220 495 792 924 792 495 220 66 12 1
 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
 1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
 1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1

ALGOL 68

Iterative - unoptimised

Translation of: C

- note: This specimen retains the original C coding style.

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d

PROC factorial = (INT n)INT:
(
        INT result;
 
        result := 1;
        FOR i  TO n DO
                result *:= i
        OD;
 
        result
);
 
PROC choose = (INT n, INT k)INT:
(
        INT result;

# Note: code can be optimised here as k < n #
        result := factorial(n) OVER (factorial(k) * factorial(n - k));
 
        result
);

test:(
        print((choose(5, 3), new line))
)

Output:

+10

ALGOL W

begin
    % calculates n!/k!                                                       %
    integer procedure factorialOverFactorial( integer value n, k ) ;
        if      k > n then 0
        else if k = n then 1
        else %  k < n %    begin
            integer f;
            f := 1;
            for i := k + 1 until n do f := f * i;
            f
        end factorialOverFactorial ;

    % calculates n!                                                          %
    integer procedure factorial( integer value n ) ;
        begin
            integer f;
            f := 1;
            for i := 2 until n do f := f * i;
            f
        end factorial ;

    % calculates the binomial coefficient of (n k)                           %
    % uses the factorialOverFactorial procedure for a slight optimisation    %
    integer procedure binomialCoefficient( integer value n, k ) ;
        if ( n - k ) > k
        then factorialOverFactorial( n, n - k ) div factorial(   k   )
        else factorialOverFactorial( n,   k   ) div factorial( n - k );

    % display the binomial coefficient of (5 3)                              %
    write( binomialCoefficient( 5, 3 ) )

end.

APL

When the factorial operator ! is used as a dyad, it returns the binomial coefficient: k!n = n choose k.

      3!5  
10

AppleScript

Imperative

set n to 5
set k to 3

on calculateFactorial(val)
	set partial_factorial to 1 as integer
	repeat with i from 1 to val
		set factorial to i * partial_factorial
		set partial_factorial to factorial
	end repeat
	return factorial
end calculateFactorial

set n_factorial to calculateFactorial(n)
set k_factorial to calculateFactorial(k)
set n_minus_k_factorial to calculateFactorial(n - k)

return n_factorial / (n_minus_k_factorial) * 1 / (k_factorial) as integer

Functional

Using a little more abstraction for readability, and currying for ease of both re-use and refactoring:

-- factorial :: Int -> Int
on factorial(n)
    product(enumFromTo(1, n))
end factorial


-- binomialCoefficient :: Int -> Int -> Int
on binomialCoefficient(n, k)
    factorial(n) div (factorial(n - k) * (factorial(k)))
end binomialCoefficient

-- Or, by reduction:

-- binomialCoefficient2 :: Int -> Int -> Int
on binomialCoefficient2(n, k)
    product(enumFromTo(1 + k, n)) div (factorial(n - k))
end binomialCoefficient2


-- TEST -----------------------------------------------------
on run
    
    {binomialCoefficient(5, 3), binomialCoefficient2(5, 3)}
    
    --> {10, 10}
end run


-- GENERAL -------------------------------------------------

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


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

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

-- product :: [Num] -> Num
on product(xs)
    script multiply
        on |λ|(a, b)
            a * b
        end |λ|
    end script
    
    foldl(multiply, 1, xs)
end product

Output:

{10, 10}

AWK

# syntax: GAWK -f EVALUATE_BINOMIAL_COEFFICIENTS.AWK
BEGIN {
    main(5,3)
    main(100,2)
    main(33,17)
    exit(0)
}
function main(n,k,  i,r) {
    r = 1
    for (i=1; i<k+1; i++) {
      r *= (n - i + 1) / i
    }
    printf("%d %d = %d\n",n,k,r)
}

Output:

5 3 = 10
100 2 = 4950
33 17 = 1166803110

Batch File

@echo off & setlocal

if "%~2"=="" ( echo Usage: %~nx0 n k && goto :EOF )

call :binom binom %~1 %~2
1>&2 set /P "=%~1 choose %~2 = "<NUL
echo %binom%

goto :EOF

:binom <var_to_set> <N> <K>
setlocal
set /a coeff=1, nk=%~2 - %~3 + 1
for /L %%I in (%nk%, 1, %~2) do set /a coeff *= %%I
for /L %%I in (1, 1, %~3) do set /a coeff /= %%I
endlocal && set "%~1=%coeff%"
goto :EOF

Output:

> binom.bat 5 3
5 choose 3 = 10

> binom.bat 100 2
100 choose 2 = 4950

The string n choose k = is output to stderr, while the result is echoed to stdout. This should allow capturing the result with a for /f loop without needing to define tokens or delims.

But...

> binom.bat 33 17
33 choose 17 = 0

> binom.bat 15 10
15 choose 10 = -547

The Windows cmd console only handles 32-bit integers. If a factoral exceeds 2147483647 at any point, set /a will choke and roll over to a negative value, giving unexpected results. Unfortunately, this is as good as it gets for pure batch.

BCPL

GET "libhdr"

LET choose(n, k) =
	~(0 <= k <= n) -> 0,
	2*k > n -> binomial(n, n - k),
	binomial(n, k)

AND binomial(n, k) =
	k = 0 -> 1,
	binomial(n, k - 1) * (n - k + 1) / k

LET start() = VALOF {
    LET n, k = ?, ?
    LET argv = VEC 20
    LET sz = ?

    sz := rdargs("n/a/n/p,k/a/n/p", argv, 20)
    UNLESS sz ~= 0 RESULTIS 1

    n := !argv!0
    k := !argv!1

    writef("%d  choose %d  = %d *n", n, k, choose(n, k))
    RESULTIS 0
}

Output:

Note that with the /p flag to rdargs(), the system will prompt if we don't supply both arguments on the command line.

$ cintsys64

BCPL 64-bit Cintcode System (13 Jan 2020)
0.004> nCk 50 25
50 choose 25 = 126410606437752
0.003> nCk 10 5
10 choose 5 = 252
0.004> nCk 100 2
100 choose 2 = 4950
0.000> nCk 100 98
100 choose 98 = 4950
0.004> nCk
n > 5
k > 3
5 choose 3 = 10

BBC BASIC

      @%=&1010
      
      PRINT "Binomial (5,3) = "; FNbinomial(5, 3)
      PRINT "Binomial (100,2) = "; FNbinomial(100, 2)
      PRINT "Binomial (33,17) = "; FNbinomial(33, 17)
      END
      
      DEF FNbinomial(N%, K%)
      LOCAL R%, D%
      R% = 1 : D% = N% - K%
      IF D% > K% THEN K% = D% : D% = N% - K%
      WHILE N% > K%
        R% *= N%
        N% -= 1
        WHILE D% > 1 AND (R% MOD D%) = 0
          R% /= D%
          D% -= 1
        ENDWHILE
      ENDWHILE
      = R%

Output:

Binomial (5,3) = 10
Binomial (100,2) = 4950
Binomial (33,17) = 1166803110

Bracmat

(binomial=
  n k coef
.   !arg:(?n,?k)
  & (!n+-1*!k:<!k:?k|)
  & 1:?coef
  &   whl
    ' ( !k:>0
      & !coef*!n*!k^-1:?coef
      & !k+-1:?k
      & !n+-1:?n
      )
  & !coef
);

binomial$(5,3)
10

Burlesque

blsq ) 5 3nr
10

C

#include <stdio.h>
#include <limits.h>

/* We go to some effort to handle overflow situations */

static unsigned long gcd_ui(unsigned long x, unsigned long y) {
  unsigned long t;
  if (y < x) { t = x; x = y; y = t; }
  while (y > 0) {
    t = y;  y = x % y;  x = t;  /* y1 <- x0 % y0 ; x1 <- y0 */
  }
  return x;
}

unsigned long binomial(unsigned long n, unsigned long k) {
  unsigned long d, g, r = 1;
  if (k == 0) return 1;
  if (k == 1) return n;
  if (k >= n) return (k == n);
  if (k > n/2) k = n-k;
  for (d = 1; d <= k; d++) {
    if (r >= ULONG_MAX/n) {  /* Possible overflow */
      unsigned long nr, dr;  /* reduced numerator / denominator */
      g = gcd_ui(n, d);  nr = n/g;  dr = d/g;
      g = gcd_ui(r, dr);  r = r/g;  dr = dr/g;
      if (r >= ULONG_MAX/nr) return 0;  /* Unavoidable overflow */
      r *= nr;
      r /= dr;
      n--;
    } else {
      r *= n--;
      r /= d;
    }
  }
  return r;
}

int main() {
    printf("%lu\n", binomial(5, 3));
    printf("%lu\n", binomial(40, 19));
    printf("%lu\n", binomial(67, 31));
    return 0;
}

Output:

10
131282408400
11923179284862717872

C#

using System;

namespace BinomialCoefficients
{
    class Program
    {
        static void Main(string[] args)
        {
            ulong n = 1000000, k = 3;
            ulong result = biCoefficient(n, k);
            Console.WriteLine("The Binomial Coefficient of {0}, and {1}, is equal to: {2}", n, k, result);
            Console.ReadLine();
        }

        static int fact(int n)
        {
            if (n == 0) return 1;
            else return n * fact(n - 1);
        }

        static ulong biCoefficient(ulong n, ulong k)
        {
            if (k > n - k)
            {
                k = n - k;
            }

            ulong c = 1;
            for (uint i = 0; i < k; i++)
            {
                c = c * (n - i);
                c = c / (i + 1);
            }
            return c;
        }
    }
}

C++

double Factorial(double nValue)
   {
       double result = nValue;
       double result_next;
       double pc = nValue;
       do
       {
           result_next = result*(pc-1);
           result = result_next;
           pc--;
       }while(pc>2);
       nValue = result;
       return nValue;
   }

double binomialCoefficient(double n, double k)
   {
       if (abs(n - k) < 1e-7 || k  < 1e-7) return 1.0;
       if( abs(k-1.0) < 1e-7 || abs(k - (n-1)) < 1e-7)return n;
       return Factorial(n) /(Factorial(k)*Factorial((n - k)));
   }

Implementation:

int main()
{
    cout<<"The Binomial Coefficient of 5, and 3, is equal to: "<< binomialCoefficient(5,3);
    cin.get();
}

Output:

The Binomial Coefficient of 5, and 3, is equal to: 10

Clojure

(defn binomial-coefficient [n k]
  (let [rprod (fn [a b] (reduce * (range a (inc b))))]
    (/ (rprod (- n k -1) n) (rprod 1 k))))

CoffeeScript

binomial_coefficient = (n, k) ->
  result = 1
  for i in [0...k]
    result *= (n - i) / (i + 1)
  result
  
n = 5
for k in [0..n]
  console.log "binomial_coefficient(#{n}, #{k}) = #{binomial_coefficient(n,k)}"

Output:


> coffee binomial.coffee 
binomial_coefficient(5, 0) = 1
binomial_coefficient(5, 1) = 5
binomial_coefficient(5, 2) = 10
binomial_coefficient(5, 3) = 10
binomial_coefficient(5, 4) = 5
binomial_coefficient(5, 5) = 1

Commodore BASIC

10 REM BINOMIAL COEFFICIENTS
20 REM COMMODORE BASIC 2.0
30 REM 2021-08-24
40 REM BY ALVALONGO
100 Z=0:U=1
110 FOR N=U TO 10
120 PRINT N;
130 FOR K=Z TO N
140 GOSUB 900
150 PRINT C;
160 NEXT K
170 PRINT
180 NEXT N
190 END
900 REM BINOMIAL COEFFICIENT
910 IF K<Z OR K>N THEN C=Z:RETURN
920 IF K=Z OR K=N THEN C=U:RETURN
930 P=K:IF N-K<P THEN P=N-K
940 C=U
950 FOR I=Z TO P-U
960 C=C/(I+U)*(N-I)
980 NEXT I
990 RETURN

Common Lisp

(defun choose (n k)
  (labels ((prod-enum (s e)
	     (do ((i s (1+ i)) (r 1 (* i r))) ((> i e) r)))
	   (fact (n) (prod-enum 1 n)))
    (/ (prod-enum (- (1+ n) k) n) (fact k))))

D

T binomial(T)(in T n, T k) pure nothrow {
    if (k > (n / 2))
        k = n - k;
    T bc = 1;
    foreach (T i; T(2) .. k + 1)
        bc = (bc * (n - k + i)) / i;
    return bc;
}

void main() {
    import std.stdio, std.bigint;

    foreach (const d; [[5, 3], [100, 2], [100, 98]])
        writefln("(%3d %3d) = %s", d[0], d[1], binomial(d[0], d[1]));
    writeln("(100  50) = ", binomial(100.BigInt, 50.BigInt));
}

Output:

(  5   3) = 2
(100   2) = 50
(100  98) = 50
(100  50) = 1976664223067613962806675336

The above wouldn't work for me (100C50 correctly gives 100891344545564193334812497256).

Translation of: C#

T BinomialCoeff(T)(in T n, in T k)
{
    T nn = n, kk = k, c = cast(T)1;

    if (kk > nn - kk) kk = nn - kk;

    for (T i = cast(T)0; i < kk; i++)
    {
        c = c * (nn - i);
        c = c / (i + cast(T)1);
    }

    return c;
}

void main()
{
    import std.stdio, std.bigint;

    BinomialCoeff(10UL, 3UL).writeln;
    BinomialCoeff(100.BigInt, 50.BigInt).writeln;
}

Output:

120
100891344545564193334812497256

dc

[sx1q]sz[d0=zd1-lfx*]sf[skdlfxrlk-lfxlklfx*/]sb

Demonstration:

5 3lbxp

10

Annotated version:

[ macro z: factorial base case when n is (z)ero ]sx
[sx     [ x is our dump register; get rid of extraneous copy of n we no longer need]sx
 1      [ return value is 1 ]sx
 q]     [ abort processing of calling macro ]sx
sz      

[ macro f: factorial ]sx [
  d       [ duplicate the input (n) ]sx
  0 =z    [ if n is zero, call z, which stops here and returns 1 ]sx
  d       [ otherwise, duplicate n again ]sx
  1 -     [ subtract 1 ]sx
  lfx     [ take the factorial ]sx
  *       [ we have (n-1)!; multiply it by the copy of n to get n! ]sx
] sf    

[ macro b(n,k): binomial function (n choose k). 
  straightforward RPN version of formula.]sx [
  sk      [ remember k. stack:              n       ]sx
  d       [ duplicate:             n        n       ]sx
  lfx     [ call factorial:        n        n!      ]sx
  r       [ swap:                  n!       n       ]sx 
  lk      [ load k:           n!   n        k       ]sx
  -       [ subtract:              n!      n-k      ]sx
  lfx     [ call factorial:        n!     (n-k)!    ]sx
  lk      [ load k:           n! (n-k)!     k       ]sx
  lfx     [ call factorial;   n! (n-k)!     k!      ]sx
  *       [ multiply:              n!    (n-k)!k!   ]sx
  /       [ divide:                     n!/(n-k)!k! ]sx
] sb      

5 3 lb x p  [print(5 choose 3)]sx

Delphi

program Binomial;

{$APPTYPE CONSOLE}

function BinomialCoff(N, K: Cardinal): Cardinal;
var
  L: Cardinal;

begin
  if N < K then
    Result:= 0      // Error
  else begin
    if K > N - K then
      K:= N - K;    // Optimization
    Result:= 1;
    L:= 0;
    while L < K do begin
      Result:= Result * (N - L);
      Inc(L);
      Result:= Result div L;
    end;
  end;
end;

begin
  Writeln('C(5,3) is ', BinomialCoff(5, 3));
  ReadLn;
end.

EasyLang

func binomial n k .
   if k > n / 2
      k = n - k
   .
   numer = 1
   for i = n downto n - k + 1
      numer = numer * i
   .
   denom = 1
   for i = 1 to k
      denom = denom * i
   .
   return numer / denom
.
print binomial 5 3

Elixir

Translation of: Erlang

defmodule RC do
  def choose(n,k) when is_integer(n) and is_integer(k) and n>=0 and k>=0 and n>=k do
    if k==0, do: 1, else: choose(n,k,1,1)
  end
  
  def choose(n,k,k,acc), do: div(acc * (n-k+1), k)
  def choose(n,k,i,acc), do: choose(n, k, i+1, div(acc * (n-i+1), i))
end

IO.inspect RC.choose(5,3)
IO.inspect RC.choose(60,30)

Output:

10
118264581564861424

Erlang

choose(N, 0) -> 1;
choose(N, K) when is_integer(N), is_integer(K), (N >= 0), (K >= 0), (N >= K) ->
  choose(N, K, 1, 1).

choose(N, K, K, Acc) ->
  (Acc * (N-K+1)) div K;
choose(N, K, I, Acc) ->
  choose(N, K, I+1, (Acc * (N-I+1)) div I).

ERRE

PROGRAM BINOMIAL

!$DOUBLE

PROCEDURE BINOMIAL(N,K->BIN)
      LOCAL R,D
      R=1 D=N-K
      IF D>K THEN K=D D=N-K  END IF
      WHILE N>K DO
        R*=N
        N-=1
        WHILE D>1 AND (R-D*INT(R/D))=0 DO
          R/=D
          D-=1
        END WHILE
      END WHILE
      BIN=R
END PROCEDURE

BEGIN
   BINOMIAL(5,3->BIN)
   PRINT("Binomial (5,3) = ";BIN)
   BINOMIAL(100,2->BIN)
   PRINT("Binomial (100,2) = ";BIN)
   BINOMIAL(33,17->BIN)
   PRINT("Binomial (33,17) = ";BIN)
END PROGRAM

Output:

Binomial (5,3) =  10
Binomial (100,2) =  4950
Binomial (33,17) =  1166803110

F#

let choose n k = List.fold (fun s i -> s * (n-i+1)/i ) 1 [1..k]

Factor

: fact ( n -- n-factorial )
    dup 0 = [ drop 1 ] [ dup 1 - fact * ] if ;

: choose ( n k -- n-choose-k )
    2dup - [ fact ] tri@ * / ;

! outputs 10
5 3 choose .

! alternative using folds
USE: math.ranges

! (product [n..k+1] / product [n-k..1])
: choose-fold ( n k -- n-choose-k )
    2dup 1 + [a,b] product -rot - 1 [a,b] product / ;

Fermat

The binomial function is built in.

Bin(5,3)

Output:

Forth

: choose ( n k -- nCk ) 1 swap 0 ?do over i - i 1+ */ loop nip ;

 5  3 choose .   \ 10
33 17 choose .   \ 1166803110

Fortran

Direct Method

Works with: Fortran version 90 and later

program test_choose

  implicit none

  write (*, '(i0)') choose (5, 3)

contains

  function factorial (n) result (res)

    implicit none
    integer, intent (in) :: n
    integer :: res
    integer :: i

    res = product ((/(i, i = 1, n)/))

  end function factorial

  function choose (n, k) result (res)

    implicit none
    integer, intent (in) :: n
    integer, intent (in) :: k
    integer :: res

    res = factorial (n) / (factorial (k) * factorial (n - k))

  end function choose

end program test_choose

Output:

Avoiding Overflow

Of course this method doesn't avoid overflow completely just delays it. It could be extended by adding more entries to the primes array

program binomial
    integer :: i, j
    
    do j=1,20
        write(*,fmt='(i2,a)',advance='no') j,'Cr = '
        do i=0,j
            write(*,fmt='(i0,a)',advance='no') n_C_r(j,i),' '
        end do
        write(*,'(a,i0)') ' 60C30 = ',n_C_r(60,30)
    end do
    stop
    
contains

    pure function n_C_r(n, r) result(bin)
        integer(16)         :: bin
        integer, intent(in) :: n
        integer, intent(in) :: r
        
        integer(16)         :: num
        integer(16)         :: den
        integer             :: i
        integer             :: k
        integer, parameter  :: primes(*) = [2,3,5,7,11,13,17,19]
        num = 1
        den = 1
        do i=0,r-1
            num = num*(n-i)
            den = den*(i+1)
            if (i > 0) then
                ! Divide out common prime factors
                do k=1,size(primes)
                    if (mod(i,primes(k)) == 0) then
                        num = num/primes(k)
                        den = den/primes(k)
                    end if
                end do
            end if
        end do
        bin = num/den
    end function n_C_r        

end program binomial

Output:

 1Cr = 1 1 
 2Cr = 1 2 1 
 3Cr = 1 3 3 1 
 4Cr = 1 4 6 4 1 
 5Cr = 1 5 10 10 5 1 
 6Cr = 1 6 15 20 15 6 1 
 7Cr = 1 7 21 35 35 21 7 1 
 8Cr = 1 8 28 56 70 56 28 8 1 
 9Cr = 1 9 36 84 126 126 84 36 9 1 
10Cr = 1 10 45 120 210 252 210 120 45 10 1 
11Cr = 1 11 55 165 330 462 462 330 165 55 11 1 
12Cr = 1 12 66 220 495 792 924 792 495 220 66 12 1 
13Cr = 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 
14Cr = 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 
15Cr = 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1 
16Cr = 1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1 
17Cr = 1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1 
18Cr = 1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1 
19Cr = 1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1 
20Cr = 1 20 190 1140 4845 15504 38760 77520 125970 167960 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1 
21Cr = 1 21 210 1330 5985 20349 54264 116280 203490 293930 352716 352716 293930 203490 116280 54264 20349 5985 1330 210 21 1 
22Cr = 1 22 231 1540 7315 26334 74613 170544 319770 497420 646646 705432 646646 497420 319770 170544 74613 26334 7315 1540 231 22 1 
23Cr = 1 23 253 1771 8855 33649 100947 245157 490314 817190 1144066 1352078 1352078 1144066 817190 490314 245157 100947 33649 8855 1771 253 23 1 
24Cr = 1 24 276 2024 10626 42504 134596 346104 735471 1307504 1961256 2496144 2704156 2496144 1961256 1307504 735471 346104 134596 42504 10626 2024 276 24 1 
25Cr = 1 25 300 2300 12650 53130 177100 480700 1081575 2042975 3268760 4457400 5200300 5200300 4457400 3268760 2042975 1081575 480700 177100 53130 12650 2300 300 25 1
60C30 = 118264581564861424

FreeBASIC

' FB 1.05.0 Win64

Function factorial(n As Integer) As Integer
  If n < 1 Then Return 1
  Dim product As Integer = 1
  For i As Integer = 2 To n
    product *= i
  Next
  Return Product
End Function

Function binomial(n As Integer, k As Integer) As Integer
  If n < 0 OrElse k < 0 OrElse n <= k Then Return 1
  Dim product As Integer = 1
  For i As Integer = n - k + 1 To n
    Product *= i
  Next
  Return product \ factorial(k)
End Function

For n As Integer =  0 To 14 
  For k As Integer = 0 To n
    Print Using "####"; binomial(n, k); 
    Print" ";
  Next k
  Print
Next n

Print
Print "Press any key to quit"
Sleep

Output:

   1
   1    1
   1    2    1
   1    3    3    1
   1    4    6    4    1
   1    5   10   10    5    1
   1    6   15   20   15    6    1
   1    7   21   35   35   21    7    1
   1    8   28   56   70   56   28    8    1
   1    9   36   84  126  126   84   36    9    1
   1   10   45  120  210  252  210  120   45   10    1
   1   11   55  165  330  462  462  330  165   55   11    1
   1   12   66  220  495  792  924  792  495  220   66   12    1
   1   13   78  286  715 1287 1716 1716 1287  715  286   78   13    1
   1   14   91  364 1001 2002 3003 3432 3003 2002 1001  364   91   14    1

Frink

Frink has a built-in efficient function to find binomial coefficients. It produces arbitrarily-large integers.

println[binomial[5,3]]

FunL

FunL has pre-defined function choose in module integers, which is defined as:

def
  choose( n, k ) | k < 0 or k > n = 0
  choose( n, 0 ) = 1
  choose( n, n ) = 1
  choose( n, k ) = product( [(n - i)/(i + 1) | i <- 0:min( k, n - k )] )

println( choose(5, 3) )
println( choose(60, 30) )

Output:

10
118264581564861424

Here it is defined using the recommended formula for this task.

import integers.factorial

def
  binomial( n, k ) | k < 0 or k > n = 0
  binomial( n, k ) = factorial( n )/factorial( n - k )/factorial( k )

GAP

# Built-in
Binomial(5, 3);
# 10

Go

package main
import "fmt"
import "math/big"

func main() {
  fmt.Println(new(big.Int).Binomial(5, 3))
  fmt.Println(new(big.Int).Binomial(60, 30))
}

Output:

10
118264581564861424

Golfscript

Actually evaluating n!/(k! (n-k)!):

;5 3 # Set up demo input
{),(;{*}*}:f; # Define a factorial function
.f@.f@/\@-f/

But Golfscript is meant for golfing, and it's shorter to calculate $\prod _{i=0}^{k-1}{\frac {n-i}{i+1}}$ :

;5 3 # Set up demo input
1\,@{1$-@\*\)/}+/

Groovy

Solution:

def factorial = { x ->
    assert x > -1
    x == 0 ? 1 : (1..x).inject(1G) { BigInteger product, BigInteger factor -> product *= factor }
}

def combinations = { n, k ->
    assert k >= 0
    assert n >= k
    factorial(n).intdiv(factorial(k)*factorial(n-k))
}

Test:

assert combinations(20, 0) == combinations(20, 20)
assert combinations(20, 10) == (combinations(19, 9) + combinations(19, 10))
assert combinations(5, 3) == 10
println combinations(5, 3)

Output:

GW-BASIC

10 REM BINOMIAL CALCULATOR
20 INPUT "N? ", N
30 INPUT "P? ", P
40 GOSUB 70
50 PRINT C
60 END
70 C = 0
80 IF N < 0 OR P<0 OR P > N THEN RETURN
90 IF P < N\2 THEN P = N - P
100 C = 1
110 FOR I = N TO P+1 STEP -1
120 C=C*I
130 NEXT I
140 FOR I = 1 TO N-P
150 C=C/I
160 NEXT I
170 RETURN

Haskell

The only trick here is realizing that everything's going to divide nicely, so we can use div instead of (/).

choose :: (Integral a) => a -> a -> a
choose n k = product [k+1..n] `div` product [1..n-k]

> 5 `choose` 3
10

Or, generate the binomial coefficients iteratively to avoid computing with big numbers:

choose :: (Integral a) => a -> a -> a
choose n k = foldl (\z i -> (z * (n-i+1)) `div` i) 1 [1..k]

Or using "caching":

coeffs = iterate next [1] 
  where
    next ns = zipWith (+) (0:ns) $ ns ++ [0]

main = print $ coeffs !! 5 !! 3

HicEst

WRITE(Messagebox) BinomCoeff( 5, 3) ! displays 10

FUNCTION factorial( n )
   factorial = 1
   DO i = 1, n
      factorial = factorial * i
   ENDDO
END

FUNCTION BinomCoeff( n, k )
   BinomCoeff = factorial(n)/factorial(n-k)/factorial(k)
END

Icon and Unicon

link math, factors 

procedure main()
write("choose(5,3)=",binocoef(5,3))
end

Output:

choose(5,3)=10

Library: Icon Programming Library

math provides binocoef and factors provides factorial.

procedure binocoef(n, k)	#: binomial coefficient

   k := integer(k) | fail
   n := integer(n) | fail

   if (k = 0) | (n = k) then return 1

   if 0 <= k <= n then 
      return factorial(n) / (factorial(k) * factorial(n - k))
   else fail

end

procedure factorial(n)			#: return n! (n factorial)
   local i

   n := integer(n) | runerr(101, n)

   if n < 0 then fail

   i := 1

   every i *:= 1 to n

   return i

end

IS-BASIC

100 PROGRAM "Binomial.bas"
110 PRINT "Binomial (5,3) =";BINOMIAL(5,3)
120 DEF BINOMIAL(N,K)
130   LET R=1:LET D=N-K
140   IF D>K THEN LET K=D:LET D=N-K
150   DO WHILE N>K
160     LET R=R*N:LET N=N-1
170     DO WHILE D>1 AND MOD(R,D)=0
180       LET R=R/D:LET D=D-1
190     LOOP
200   LOOP
210   LET BINOMIAL=R
220 END DEF

J

Solution:
The dyadic form of the primitive ! ([Out of]) evaluates binomial coefficients directly.

Example usage:

   3 ! 5
10

Java

public class Binomial {

    // precise, but may overflow and then produce completely incorrect results
    private static long binomialInt(int n, int k) {
        if (k > n - k)
            k = n - k;

        long binom = 1;
        for (int i = 1; i <= k; i++)
            binom = binom * (n + 1 - i) / i;
        return binom;
    }

    // same as above, but with overflow check
    private static Object binomialIntReliable(int n, int k) {
        if (k > n - k)
            k = n - k;

        long binom = 1;
        for (int i = 1; i <= k; i++) {
            try {
                binom = Math.multiplyExact(binom, n + 1 - i) / i;
            } catch (ArithmeticException e) {
                return "overflow";
            }
        }
        return binom;
    }

    // using floating point arithmetic, larger numbers can be calculated,
    // but with reduced precision
    private static double binomialFloat(int n, int k) {
        if (k > n - k)
            k = n - k;

        double binom = 1.0;
        for (int i = 1; i <= k; i++)
            binom = binom * (n + 1 - i) / i;
        return binom;
    }

    // slow, hard to read, but precise
    private static BigInteger binomialBigInt(int n, int k) {
        if (k > n - k)
            k = n - k;

        BigInteger binom = BigInteger.ONE;
        for (int i = 1; i <= k; i++) {
            binom = binom.multiply(BigInteger.valueOf(n + 1 - i));
            binom = binom.divide(BigInteger.valueOf(i));
        }
        return binom;
    }

    private static void demo(int n, int k) {
        List<Object> data = Arrays.asList(
                n,
                k,
                binomialInt(n, k),
                binomialIntReliable(n, k),
                binomialFloat(n, k),
                binomialBigInt(n, k));

        System.out.println(data.stream().map(Object::toString).collect(Collectors.joining("\t")));
    }

    public static void main(String[] args) {
        demo(5, 3);
        demo(1000, 300);
    }
}

Output:

5	3	10	10	10.0	10
1000	300	-8357011479171942	overflow	5.428250046406143E263	542825004640614064815358503892902599588060075560435179852301016412253602009800031872232761420804306539976220810204913677796961128392686442868524741815732892024613137013599170443939815681313827516308854820419235457578544489551749630302863689773725905288736148678480

Recursive version, without overflow check:

public class Binomial
{
    private static long binom(int n, int k)
    {
        if (k==0)
            return 1;
        else if (k>n-k)
            return binom(n, n-k);
        else
            return binom(n-1, k-1)*n/k;
    }

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

Output:

JavaScript

function binom(n, k) {
    var coeff = 1;
    var i;

    if (k < 0 || k > n) return 0;

    for (i = 0; i < k; i++) {
        coeff = coeff * (n - i) / (i + 1);
    }
  
    return coeff;
}

console.log(binom(5, 3));

Output:

jq

# nCk assuming n >= k
def binomial(n; k): 
  if k > n / 2 then binomial(n; n-k)
  else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)
  end;

def task:
  .[0] as $n | .[1] as $k
  | "\($n) C \($k) = \(binomial( $n; $k) )";
;

([5,3], [100,2], [ 33,17]) | task

Output:

5 C 3 = 10
100 C 2 = 4950
33 C 17 = 1166803110

Julia

Works with: Julia version 1.2

Built-in

@show binomial(5, 3)

Recursive version:

function binom(n::Integer, k::Integer)
    n ≥ k || return 0 # short circuit base cases
    (n == 1 || k == 0) && return 1

    n * binom(n - 1, k - 1) ÷ k
end

@show binom(5, 3)

Output:

binomial(5, 3) = 10
binom(5, 3) = 10

K

   {[n;k]_(*/(k-1)_1+!n)%(*/1+!k)} . 5 3
10

Alternative version:

   {[n;k]i:!(k-1);_*/((n-i)%(i+1))} . 5 3
10

Using Pascal's triangle:

   pascal:{x{+':0,x,0}\1}
   pascal 5
(1
 1 1
 1 2 1
 1 3 3 1
 1 4 6 4 1
 1 5 10 10 5 1)

   {[n;k](pascal n)[n;k]} . 5 3
10

Kotlin

// version 2.0

fun binomial(n: Int, k: Int) = when {
    n < 0 || k < 0 -> throw IllegalArgumentException("negative numbers not allowed")
    n == k         -> 1L
    else           -> {
        val kReduced = min(k, n - k)    // minimize number of steps
        var result = 1L
        var numerator = n
        var denominator = 1
        while (denominator <= kReduced)
            result = result * numerator-- / denominator++
        result
    }
}

fun main(args: Array<String>) {
    for (n in 0..14) {
        for (k in 0..n)
            print("%4d ".format(binomial(n, k)))
        println()
    }
}

Output:

   1
   1    1
   1    2    1
   1    3    3    1
   1    4    6    4    1
   1    5   10   10    5    1
   1    6   15   20   15    6    1
   1    7   21   35   35   21    7    1
   1    8   28   56   70   56   28    8    1
   1    9   36   84  126  126   84   36    9    1
   1   10   45  120  210  252  210  120   45   10    1
   1   11   55  165  330  462  462  330  165   55   11    1
   1   12   66  220  495  792  924  792  495  220   66   12    1
   1   13   78  286  715 1287 1716 1716 1287  715  286   78   13    1
   1   14   91  364 1001 2002 3003 3432 3003 2002 1001  364   91   14    1

Lambdatalk

{def C
 {lambda {:n :p}
  {/ {* {S.serie :n {- :n :p -1} -1}}
     {* {S.serie :p 1 -1}}}}}
-> C

{C 16 8}
-> 12870

1{S.map {lambda {:n} {br}1 
 {S.map {C :n} {S.serie 1 {- :n 1}}} 1}
               {S.serie 2 16}}
->
1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1

Lasso

define binomial(n::integer,k::integer) => {
	#k == 0 ? return 1
	local(result = 1)
	loop(#k) => {
		#result = #result * (#n - loop_count + 1) / loop_count
	}
	return #result
}
// Tests
binomial(5, 3)
binomial(5, 4)
binomial(60, 30)

Output:

10
5
118264581564861424

Liberty BASIC

    '   [RC] Binomial Coefficients

    print "Binomial Coefficient of "; 5; " and "; 3; " is ",BinomialCoefficient( 5, 3)
    n =1 +int( 10 *rnd( 1))
    k =1 +int( n *rnd( 1))
    print "Binomial Coefficient of "; n; " and "; k; " is ",BinomialCoefficient( n, k)

    end

    function BinomialCoefficient( n, k)
        BinomialCoefficient =factorial( n) /factorial( n -k) /factorial( k)
    end function

    function factorial( n)
        if n <2 then
            f =1
        else
            f =n *factorial( n -1)
        end if
    factorial =f
    end function

Logo

to choose :n :k
  if :k = 0 [output 1]
  output (choose :n :k-1) * (:n - :k + 1) / :k
end

show choose 5 3   ; 10
show choose 60 30 ; 1.18264581564861e+17

Lua

function Binomial( n, k )
    if k > n then return nil end
    if k > n/2 then k = n - k end       --   (n k) = (n n-k)
    
    numer, denom = 1, 1
    for i = 1, k do
        numer = numer * ( n - i + 1 )
        denom = denom * i
    end
    return numer / denom
end

Additive recursion with memoization by hashing 2 input integer. Lua 5.3 support bit-wise operation; assume 64 bit integer implementation here.

local Binomial = setmetatable({},{
 __call = function(self,n,k)
   local hash = (n<<32) | (k & 0xffffffff)
   local ans = self[hash]
   if not ans then 
    if n<0 or k>n then
      return 0 -- not save
    elseif n<=1 or k==0 or k==n then 
      ans = 1
    else
      if 2*k > n then 
        ans = self(n, n - k) 
      else
        local lhs = self(n-1,k)
        local rhs = self(n-1,k-1)        
        local sum = lhs + rhs
        if sum<0 or not math.tointeger(sum)then 
          -- switch to double
          ans = lhs/1.0 + rhs/1.0 -- approximate
        else
          ans = sum
        end
      end
    end
    rawset(self,hash,ans)
   end
   return ans
 end 
})
print( Binomial(100,50)) -- 1.0089134454556e+029

Maple

convert(binomial(n,k),factorial);

binomial(5,3);

Output:

                         factorial(n)         
                 -----------------------------
                 factorial(k) factorial(n - k)

                               10

Mathematica / Wolfram Language

(Local) In[1]:= Binomial[5,3]
(Local) Out[1]= 10

MATLAB / Octave

This is a built-in function in MATLAB called "nchoosek(n,k)". But, this will only work for scalar inputs. If "n" is a vector then "nchoosek(v,k)" finds all combinations of choosing "k" elements out of the "v" vector (see Combinations#MATLAB).

Solution:

>> nchoosek(5,3)
ans =
    10

Alternative implementations are:

function r = binomcoeff1(n,k)
    r = diag(rot90(pascal(n+1))); % vector of all binomial coefficients for order n
    r = r(k); 
end;

function r = binomcoeff2(n,k)
   prod((n-k+1:n)./(1:k))
end;

function r = binomcoeff3(n,k)
   m = pascal(max(n-k,k)+1); 
   r = m(n-k+1,k+1);
end;

If you want a vectorized function that returns multiple binomial coefficients given vector inputs, you must define that function yourself. A sample implementation is given below. This function takes either scalar or vector inputs for "n" and "v" and returns either a: scalar, vector, or matrix. Where the columns are indexed by the "k" vector and the rows indexed by the "n" vector. binomialCoeff.m:

function coefficients = binomialCoeff(n,k)

    coefficients = zeros(numel(n),numel(k)); %Preallocate memory

    columns = (1:numel(k)); %Preallocate row and column counters
    rows = (1:numel(n));
    
    %Iterate over every row and column. The rows represent the n number,
    %and the columns represent the k number. If n is ever greater than k,
    %the nchoosek function will throw an error. So, we test to make sure
    %it isn't, if it is then we leave that entry in the coefficients matrix
    %zero. Which makes sense combinatorically.
    for row = rows
        for col = columns
            if k(col) <= n(row)
                coefficients(row,col) = nchoosek(n(row),k(col));
            end
        end
    end
    
end %binomialCoeff

Sample Usage:

>> binomialCoeff((0:5),(0:5))

ans =

     1     0     0     0     0     0
     1     1     0     0     0     0
     1     2     1     0     0     0
     1     3     3     1     0     0
     1     4     6     4     1     0
     1     5    10    10     5     1

>> binomialCoeff([1 0 3 2],(0:3))

ans =

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

>> binomialCoeff(3,(0:3))

ans =

     1     3     3     1

>> binomialCoeff((0:3),2)

ans =

     0
     0
     1
     3

>> binomialCoeff(5,3)

ans =

    10

Maxima

binomial( 5,  3);      /* 10 */
binomial(-5,  3);      /* -35 */
binomial( 5, -3);      /* 0 */
binomial(-5, -3);      /* 0 */
binomial( 3,  5);      /* 0 */

binomial(x, 3);        /* ((x - 2)*(x - 1)*x)/6 */

binomial(3, 1/2);      /* binomial(3, 1/2) */
makegamma(%);          /* 32/(5*%pi) */

binomial(a, b);        /* binomial(a, b) */
makegamma(%);          /* gamma(a + 1)/(gamma(-b + a + 1)*gamma(b + 1)) */

min

Works with: min version 0.19.3

((dup 0 ==) 'succ (dup pred) '* linrec) :fact
('dup dip dup ((fact) () (- fact) (fact * div)) spread) :binomial

5 3 binomial puts!

Output:

MINIL

// Number of combinations nCr
00 0E  Go:    ENT  R0   // n
01 1E         ENT  R1   // r
02 2C         CLR  R2
03 2A  Loop:  ADD1 R2
04 0D         DEC  R0
05 1D         DEC  R1
06 C3         JNZ  Loop
07 3C         CLR  R3   // for result
08 3A         ADD1 R3
09 0A  Next:  ADD1 R0
0A 1A         ADD1 R1
0B 50         R5 = R0
0C 5D         DEC  R5
0D 63         R6 = R3
0E 46  Mult:  R4 = R6
0F 3A  Add:   ADD1 R3
10 4D         DEC  R4
11 CF         JNZ  Add
12 5D         DEC  R5
13 CE         JNZ  Mult
14 61  Divide:R6 = R1
15 5A         ADD1 R5
16 3D  Sub:   DEC  R3
17 9B         JZ   Exact
18 6D         DEC  R6
19 D6         JNZ  Sub
1A 94         JZ   Divide
1B 35  Exact: R3 = R5
1C 2D         DEC  R2
1D C9         JNZ  Next
1E 03         R0 = R3
1F 80         JZ   Go   // Display result

This uses the recursive definition:

ncr(n, r) = 1 if r = 0

ncr(n, r) = n/r * ncr(n-1, r-1) otherwise

which results from the definition of ncr in terms of factorials.

МК-61/52

П1	<->	П0	ПП	22	П2	 ИП1	ПП	22	П3
ИП0	ИП1	-	ПП	22	ИП3	*	П3	ИП2	ИП3
/	С/П	ВП	П0	1	ИП0	*	L0	25	В/О

Input: n ^ k В/О С/П.

Nanoquery

Translation of: Python

def binomialCoeff(n, k)
    result = 1
    for i in range(1, k)
        result = result * (n-i+1) / i
    end
    return result
end

if main
    println binomialCoeff(5,3)
end

Nim

Note that a function to compute these coefficients, named “binom”, is available in standard module “math”.

proc binomialCoeff(n, k: int): int =
  result = 1
  for i in 1..k:
    result = result * (n-i+1) div i

echo binomialCoeff(5, 3)

Output:

Oberon-2

Works with: oo2c

MODULE Binomial;
IMPORT
  Out;

PROCEDURE For*(n,k: LONGINT): LONGINT;
VAR
  i,m,r: LONGINT;
 
BEGIN
  ASSERT(n > k);
  r := 1;
  IF k > n DIV 2 THEN m := n - k ELSE m := k END;
  FOR i := 1 TO m DO
    r := r * (n - m + i) DIV i
  END;
  RETURN r
END For;

BEGIN
  Out.Int(For(5,2),0);Out.Ln
END Binomial.

Output:

OCaml

let binomialCoeff n p =
  let p = if p < n -. p then p else n -. p in
  let rec cm res num denum =
    (* this method partially prevents overflow.
     * float type is choosen to have increased domain on 32-bits computer,
     * however algorithm ensures an integral result as long as it is possible
     *)
    if denum <= p then cm ((res *. num) /. denum) (num -. 1.) (denum +. 1.)
    else res in
  cm 1. n 1.

Alternate version using big integers

#load "nums.cma";;
open Num;;

let binomial n p =
   let m = min p (n - p) in
   if m < 0 then Int 0 else
   let rec a j v =
      if j = m then v
      else a (succ j) ((v */ (Int (n - j))) // (Int (succ j)))
   in a 0 (Int 1)
;;

Simple recursive version

open Num;;
let rec binomial n k = if n = k then Int 1 else ((binomial (n-1) k) */ Int n) // Int (n-k)

Oforth

: binomial(n, k)  | i |  1 k loop: i [ n i - 1+ * i / ] ;

Output:

>5 3 binomial .
10

Oz

Translation of: Python

declare
  fun {BinomialCoeff N K}
     {List.foldL {List.number 1 K 1}
      fun {$ Z I}
         Z * (N-I+1) div I
      end
      1}
  end
in
  {Show {BinomialCoeff 5 3}}

PARI/GP

binomial(5,3)

Pascal

See Delphi

Perl

sub binomial {
    use bigint;
    my ($r, $n, $k) = (1, @_);
    for (1 .. $k) { $r *= $n--; $r /= $_ }
    $r;
}
 
print binomial(5, 3);

Output:

Since the bigint module already has a binomial method, this could also be written as:

sub binomial {
    use bigint;
    my($n,$k) = @_;
    (0+$n)->bnok($k);
}

For better performance, especially with large inputs, one can also use something like:

Library: ntheory

use ntheory qw/binomial/;
print length(binomial(100000,50000)), "\n";

Output:

The Math::Pari module also has binomial, but it needs large amounts of added stack space for large arguments (this is due to using a very old version of the underlying Pari library).

Phix

There is a builtin choose() function which does this. From builtins/factorial.e (an autoinclude):

global function choose(integer n, k)
    atom res = 1
    for i=1 to k do
        res = (res*(n-i+1))/i
    end for
    return res
end function

Example:

?choose(5,3)

Output:

However errors will creep in should any result or interim value exceed 9,007,199,254,740,992 (on 32-bit), so:

Library: Phix/mpfr

with javascript_semantics 
include builtins\mpfr.e
sequence tests = {{5,3},{100,50},{60,30},{1200,120}}
mpz r = mpz_init()
for i=1 to length(tests) do
    integer {n,k} = tests[i]
    mpz_bin_uiui(r,n,k)
    ?mpz_get_str(r)
end for

Output:

"10"
"100891344545564193334812497256"
"118264581564861424"
"1004576581793084916475353119318331966507299414258370667602185866686463289093457468590558508056798211449853806741873396451735444387513582540860551330127062642417424083600"

Note that I have re-implemented mpz_bin_uiui() mainly for the benefit of pwa/p2js, and some tweaks may be in order (please ask if needed) to match gmp proper for -ve n.

PHP

<?php
$n=5;
$k=3;
function factorial($val){
    for($f=2;$val-1>1;$f*=$val--);
    return $f;
}
$binomial_coefficient=factorial($n)/(factorial($k)*factorial($n-$k));
echo $binomial_coefficient;
?>

Alternative version, not based on factorial

function binomial_coefficient($n, $k) {
  if ($k == 0) return 1;
  $result = 1;
  foreach (range(0, $k - 1) as $i) {
    $result *= ($n - $i) / ($i + 1);
  }
  return $result;
}

Picat

Iterative

binomial_it(N,K) = Res =>
   if K < 0 ; K > N then
     R = 0
   else
     R = 1,
     foreach(I in 0..K-1) 
       R := R * (N-I) // (I+1)
     end
   end,
   Res = R.

Using built-in factorial/1

binomial_fac(N,K) = factorial(N) // factorial(K) // factorial(N-K).

Recursion (tabled)

table
binomial_rec(_N, 0) = 1.
binomial_rec(0, _K) = 0.
binomial_rec(N, K)  = binomial_rec(N-1,K-1) + binomial_rec(N-1,K).

Test

go =>
   Tests = [[10,3],[60,30],[100,50],[400,200]],
   foreach([N,K] in Tests)
     println([N,K,binomial_it(N,K)])
   end,
   nl.

All methods prints the same result.

Output:

[10,3,120]
[60,30,118264581564861424]
[100,50,100891344545564193334812497256]
[400,200,102952500135414432972975880320401986757210925381077648234849059575923332372651958598336595518976492951564048597506774120]

binomial_rec/2 is a little slower than the two other (0.036s vs 0.002s on these tests).

PicoLisp

(de binomial (N K)
   (let f
      '((N)
         (if (=0 N) 1 (apply * (range 1 N))) )
      (/
         (f N)
         (* (f (- N K)) (f K)) ) ) )

Output:

: (binomial 5 3)
-> 10

PL/I

binomial_coefficients:
   procedure options (main);
      declare (n, k) fixed;

   get (n, k);
   put (coefficient(n, k));

coefficient: procedure (n, k) returns (fixed decimal (15));
   declare (n, k) fixed;
   return (fact(n)/ (fact(n-k) * fact(k)) );
end coefficient;

fact: procedure (n) returns (fixed decimal (15));
   declare n fixed;
   declare i fixed, f fixed decimal (15);
   f = 1;
   do i = 1 to n;
      f = f * i;
   end;
   return (f);
end fact;
end binomial_coefficients;

Output:

PowerShell

function choose($n,$k) {
    if($k -le $n -and 0 -le $k) {
        $numerator = $denominator = 1
        0..($k-1) | foreach{
            $numerator *= ($n-$_)
            $denominator *= ($_ + 1)
        }
        $numerator/$denominator
    } else {
        "$k is greater than $n or lower than 0"
    }
}
choose 5 3
choose 2 1
choose 10 10
choose 10 2
choose 10 8

Output:

PureBasic

Procedure Factor(n)
  Protected Result=1
  While n>0
    Result*n
    n-1
  Wend
  ProcedureReturn Result
EndProcedure

Macro C(n,k)
  (Factor(n)/(Factor(k)*factor(n-k)))
EndMacro

If OpenConsole()
  Print("Enter value n: "): n=Val(Input())
  Print("Enter value k: "): k=Val(Input())
  PrintN("C(n,k)= "+str(C(n,k)))
  
  Print("Press ENTER to quit"): Input()
  CloseConsole()
EndIf

Example

Enter value n: 5
Enter value k: 3
C(n,k)= 10

Python

Imperative

def binomialCoeff(n, k):
    result = 1
    for i in range(1, k+1):
        result = result * (n-i+1) / i
    return result

if __name__ == "__main__":
    print(binomialCoeff(5, 3))

Output:

Functional

from operator import mul
from functools import reduce


def comb(n,r):
    ''' calculate nCr - the binomial coefficient
    >>> comb(3,2)
    3
    >>> comb(9,4)
    126
    >>> comb(9,6)
    84
    >>> comb(20,14)
    38760
    '''
 
    if r > n-r:
        # r = n-r   for smaller intermediate values during computation
        return ( reduce( mul, range((n - (n-r) + 1), n + 1), 1)
                 // reduce( mul, range(1, (n-r) + 1), 1) )
    else:
        return ( reduce( mul, range((n - r + 1), n + 1), 1)
                 // reduce( mul, range(1, r + 1), 1) )

Or, abstracting a little more for legibility and ease of reuse, while currying for ease of mapping and general composition:

Works with: Python version 3.7

'''Evaluation of binomial coefficients'''

from functools import reduce


# binomialCoefficient :: Int -> Int -> Int
def binomialCoefficient(n):
    '''n choose k, expressed in terms of
       product and factorial functions.
    '''
    return lambda k: product(
        enumFromTo(1 + k)(n)
    ) // factorial(n - k)


# TEST ----------------------------------------------------
# main :: IO()
def main():
    '''Tests'''

    print(
        binomialCoefficient(5)(3)
    )

    # k=0 to k=5, where n=5
    print(
        list(map(
            binomialCoefficient(5),
            enumFromTo(0)(5)
        ))
    )


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

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


# factorial :: Int -> Int
def factorial(x):
    '''The factorial of x, where
       x is a positive integer.
    '''
    return product(enumFromTo(1)(x))


# product :: [Num] -> Num
def product(xs):
    '''The product of a list of
       numeric values.
    '''
    return reduce(lambda a, b: a * b, xs, 1)


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

Output:

10
[1, 5, 10, 10, 5, 1]

Compare the use of Python comments, (above); with the use of Python type hints, (below).

from typing import (Callable, List, Any)
from functools import reduce
from operator import mul


def binomialCoefficient(n: int) -> Callable[[int], int]:
    return lambda k: product(enumFromTo(1 + k)(n)) // factorial(n - k)


def enumFromTo(m: int) -> Callable[[int], List[Any]]:
    return lambda n: list(range(m, 1 + n))


def factorial(x: int) -> int:
    return product(enumFromTo(1)(x))


def product(xs: List[Any]) -> int:
    return reduce(mul, xs, 1)


if __name__ == '__main__':
    print(binomialCoefficient(5)(3))
    # k=0 to k=5, where n=5
    print(list(map(binomialCoefficient(5), enumFromTo(0)(5))))

Output:

10
[1, 5, 10, 10, 5, 1]

Quackery

  [ tuck - over
   1 swap times 
     [ over i + 1+ * ]
   nip swap times
      [ i 1+ / ] ]     is binomial ( n n --> )

  5 3 binomial echo

Output:

R

R's built-in choose() function evaluates binomial coefficients:

choose(5,3)

Output:

[1] 10

Racket

#lang racket
(require math)
(binomial 10 5)

Raku

(formerly Perl 6) For a start, you can get the length of the corresponding list of combinations:

say combinations(5, 3).elems;

Output:

This method is efficient, as Raku will not actually compute each element of the list, since it actually uses an iterator with a defined count-only method. Such method performs computations in a way similar to the following infix operator:

sub infix:<choose> { [*] ($^n ... 0) Z/ 1 .. $^p }
say 5 choose 3;

A possible optimization would use a symmetry property of the binomial coefficient:

sub infix:<choose> { [*] ($^n ... 0) Z/ 1 .. min($n - $^p, $p) }

One drawback of this method is that it returns a Rat, not an Int. So we actually may want to enforce the conversion:

sub infix:<choose> { ([*] ($^n ... 0) Z/ 1 .. min($n - $^p, $p)).Int }

And this is exactly what the count-only method does.

REXX

The task is to compute ANY binomial coefficient(s), but these REXX examples are limited to 100k digits.

idiomatic

/*REXX program calculates   binomial coefficients  (also known as  combinations).       */
numeric digits 100000                            /*be able to handle gihugeic numbers.  */
parse arg n k .                                  /*obtain  N  and  K   from the C.L.    */
say 'combinations('n","k')='  comb(n,k)          /*display the number of combinations.  */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure;  parse arg x,y;         return !(x) % (!(x-y) * !(y))
!:    procedure;  !=1;           do j=2  to arg(1);  !=!*j;  end  /*j*/;          return !

output when using the input of: 5 3

combinations(5,3)= 10

output when using the input of: 1200 120

combinations(1200,120)= 1004576581793084916475353119318331966507299414258370667602185866686463289093457468590558508056798211449853806741873396451735444387513582540860551330127062642417424083600

optimized

This REXX version takes advantage of reducing the size (product) of the numerator, and also,
only two (factorial) products need be calculated.

/*REXX program calculates   binomial coefficients  (also known as  combinations).       */
numeric digits 100000                            /*be able to handle gihugeic numbers.  */
parse arg n k .                                  /*obtain  N  and  K   from the C.L.    */
say 'combinations('n","k')='  comb(n,k)          /*display the number of combinations.  */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb:  procedure;  parse arg x,y;        return pfact(x-y+1, x)  %  pfact(2, y)
/*──────────────────────────────────────────────────────────────────────────────────────*/
pfact: procedure;  !=1;        do j=arg(1)  to arg(2);  !=!*j;  end  /*j*/;       return !

output is identical to the 1^st REXX version.

It is (around average) about ten times faster than the 1^st version for 200,20 and 100,10.
For 100,80 it is about 30% faster.

Ring

numer = 0
binomial(5,3)
see "(5,3) binomial = " + numer + nl

func binomial n, k 
     if k > n return nil ok
     if k > n/2 k = n - k ok
     numer = 1
     for i = 1 to k  
         numer = numer * ( n - i + 1 ) / i
     next 
     return numer

RPL

RPL has a COMB instruction which returns the result directly. If a home-made function is preferred, there are many possibilities.

Using the recommended formula and the stack:

≪ - LAST ! SWAP ! SWAP ROT ! * /  ≫ ‘CHOOS’ STO

Using the formula with local variables:

 ≪ → n k ≪ n ! n k - ! / k ! / ≫ ≫ ‘CHOOS’ STO

To avoid data overflow for high values of n, using the formula simplified by (n-k)! :

 ≪ → n k ≪ 1 n k - 1 + n FOR j j * NEXT k ! / ≫ ≫ ‘CHOOS’ STO

All the above functions are called the same way and return the same result:

5 3 CHOOS

Output:

Ruby

Translation of: Tcl

Works with: Ruby version 1.8.7+

class Integer
  # binomial coefficient: n C k
  def choose(k)
    # n!/(n-k)!
    pTop = (self-k+1 .. self).inject(1, &:*) 
    # k!
    pBottom = (2 .. k).inject(1, &:*)
    pTop / pBottom
  end
end

p 5.choose(3)
p 60.choose(30)

result

10
118264581564861424

another implementation:

def c n, r
  (0...r).inject(1) do |m,i| (m * (n - i)) / (i + 1) end
end

Ruby's Arrays have a combination method which result in a (lazy) enumerator. This Enumerator has a "size" method, which returns the size of the enumerator, or nil if it can’t be calculated lazily. (Since Ruby 2.0)

(1..60).to_a.combination(30).size  #=> 118264581564861424

Run BASIC

print "binomial (5,1) = "; binomial(5, 1)
print "binomial (5,2) = "; binomial(5, 2)
print "binomial (5,3) = "; binomial(5, 3)
print "binomial (5,4) = "; binomial(5,4)
print "binomial (5,5) = "; binomial(5,5)
end
 
function binomial(n,k)
 coeff = 1
 for i = n - k + 1 to n
   coeff = coeff * i
 next i
 for i = 1 to k 
   coeff = coeff / i
 next i
binomial = coeff
end function

Output:

binomial (5,1) = 5
binomial (5,2) = 10
binomial (5,3) = 10
binomial (5,4) = 5
binomial (5,5) = 1

Rust

fn fact(n:u32) -> u64 {
  let mut f:u64 = n as u64;
  for i in 2..n {
    f *= i as u64;
  }
  return f;
}

fn choose(n: u32, k: u32)  -> u64 {
   let mut num:u64 = n as u64;
   for i in 1..k {
     num *= (n-i) as u64;
   }
   return num / fact(k);
}

fn main() {
  println!("{}", choose(5,3));
}

Output:

Alternative version, using functional style:

fn choose(n:u64,k:u64)->u64 {
   let factorial=|x| (1..=x).fold(1, |a, x| a * x);
   factorial(n) / factorial(k) / factorial(n - k)
}

Scala

object Binomial {
   def main(args: Array[String]): Unit = {
      val n=5
      val k=3
      val result=binomialCoefficient(n,k)
      println("The Binomial Coefficient of %d and %d equals %d.".format(n, k, result))
   }

   def binomialCoefficient(n:Int, k:Int)=fact(n) / (fact(k) * fact(n-k))
   def fact(n:Int):Int=if (n==0) 1 else n*fact(n-1)
}

Output:

The Binomial Coefficient of 5 and 3 equals 10.

Another (more flexible and efficient) implementation. n and k are taken from command line. The use of BigInts allows to compute coefficients of arbitrary size:

object Binomial extends App {
  def binomialCoefficient(n: Int, k: Int) = 
    (BigInt(n - k + 1) to n).product / 
    (BigInt(1) to k).product

  val Array(n, k) = args.map(_.toInt)
  println("The Binomial Coefficient of %d and %d equals %,3d.".format(n, k, binomialCoefficient(n, k)))
}

Output:

java Binomial 100 30
The Binomial Coefficient of 100 and 30 equals 29,372,339,821,610,944,823,963,760.

Using recursive formula C(n,k) = C(n-1,k-1) + C(n-1,k):

  def bico(n: Long, k: Long): Long = (n, k) match {
    case (n, 0) => 1
    case (0, k) => 0
    case (n, k) => bico(n - 1, k - 1) + bico(n - 1, k)
  }
  println("bico(5,3) = " + bico(5, 3))

Output:

bico(5,3) = 10

Scheme

Works with: Scheme version R

^{5}

RS

(define (factorial n)
  (define (*factorial n acc)
    (if (zero? n)
        acc
        (*factorial (- n 1) (* acc n))))
  (*factorial n 1))

(define (choose n k)
  (/ (factorial n) (* (factorial k) (factorial (- n k)))))

(display (choose 5 3))
(newline)

Output:

Alternatively a recursive implementation can be constructed from Pascal's Triangle:

(define (pascal i j)
  (cond ((= i 0) 1)
        ((= j 0) 1)
        (else (+
               (pascal (- i 1) j)
               (pascal i (- j 1))))))

(define (choose n k)
  (pascal (- n k) k)))

(display (choose 5 3))
(newline)

Output:

Seed7

The infix operator ! computes the binomial coefficient. E.g.: 5 ! 3 evaluates to 10. The binomial coefficient operator works also for negative values of n. E.g.: (-6) ! 10 evaluates to 3003.

$ include "seed7_05.s7i";

const proc: main is func
  local
    var integer: n is 0;
    var integer: k is 0;
  begin
    for n range 0 to 66 do
      for k range 0 to n do
         write(n ! k <& " ");
      end for;
      writeln;
    end for;
  end func;

Output:

1 
1 1 
1 2 1 
1 3 3 1 
1 4 6 4 1 
1 5 10 10 5 1 
1 6 15 20 15 6 1 
1 7 21 35 35 21 7 1 
1 8 28 56 70 56 28 8 1 
1 9 36 84 126 126 84 36 9 1 
1 10 45 120 210 252 210 120 45 10 1 
1 11 55 165 330 462 462 330 165 55 11 1 
1 12 66 220 495 792 924 792 495 220 66 12 1 
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1 
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1 
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1 
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1 
1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1 
1 20 190 1140 4845 15504 38760 77520 125970 167960 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1 
...

The library bigint.s7i contains a definition of the binomial coefficient operator ! for the type bigInteger:

const func bigInteger: (in bigInteger: n) ! (in var bigInteger: k) is func
  result
    var bigInteger: binom is 0_;
  local
    var bigInteger: numerator is 0_;
    var bigInteger: denominator is 0_;
  begin
    if n >= 0_ and k > n >> 1 then
      k := n - k;
    end if;
    if k < 0_ then
      binom := 0_;
    elsif k = 0_ then
      binom := 1_;
    else
      binom := n;
      numerator := pred(n);
      denominator := 2_;
      while denominator <= k do
        binom *:= numerator;
        binom := binom div denominator;
        decr(numerator);
        incr(denominator);
      end while;
    end if;
  end func;

Original source [1].

SequenceL

Simplest Solution:

choose(n, k) := product(k + 1 ... n) / product(1 ... n - k);

Tail-Recursive solution to avoid arithmetic with large integers:

choose(n,k) := binomial(n, k, 1, 1);

binomial(n, k, i, result) :=
	result when i > k else
	binomial(n, k, i + 1, (result * (n - i + 1)) / i);

Sidef

Straightforward translation of the formula:

func binomial(n,k) {
    n! / ((n-k)! * k!)
}

say binomial(400, 200)

Alternatively, by using the Number.nok() method:

say 400.nok(200)

Smalltalk

Works with: Smalltalk/X

Having a small language but a big class library in my bag, we can write:

Transcript showCR: (5 binco:3).
Transcript showCR: (400 binco:200)

Output:

10
102952500135414432972975880320401986757210925381077648234849059575923332372651958598336595518976492951564048597506774120

A naïve implementation (in the Integer class) might look like:

binco:arg
    ^ (self factorial) / (arg factorial * (self-arg) factorial)

Standard ML

fun binomial n k =
    if k > n then 0 else
    let fun f (_, 0) = 1
          | f (i, d) = f (i + 1, d - 1) * i div d
    in f (n - k + 1, k) end

Stata

Use the comb function. Notice the result is a missing value if k>n or k<0.

. display comb(5,3)
10

Swift

func factorial<T: BinaryInteger>(_ n: T) -> T {
  guard n != 0 else {
    return 1
  }

  return stride(from: n, to: 0, by: -1).reduce(1, *)
}

func binomial<T: BinaryInteger>(_ x: (n: T, k: T)) -> T {
  let nFac = factorial(x.n)
  let kFac = factorial(x.k)

  return nFac / (factorial(x.n - x.k) * kFac)
}

print("binomial(\(5), \(3)) = \(binomial((5, 3)))")
print("binomial(\(20), \(11)) = \(binomial((20, 11)))")

Output:

binomial(5, 3) = 10
binomial(20, 11) = 167960

Tcl

This uses exact arbitrary precision integer arithmetic.

package require Tcl 8.5
proc binom {n k} {
    # Compute the top half of the division; this is n!/(n-k)!
    set pTop 1
    for {set i $n} {$i > $n - $k} {incr i -1} {
	set pTop [expr {$pTop * $i}]
    }

    # Compute the bottom half of the division; this is k!
    set pBottom 1
    for {set i $k} {$i > 1} {incr i -1} {
	set pBottom [expr {$pBottom * $i}]
    }

    # Integer arithmetic divide is correct here; the factors always cancel out
    return [expr {$pTop / $pBottom}]
}

Demonstrating:

puts "5_C_3 = [binom 5 3]"
puts "60_C_30 = [binom 60 30]"

Output:

5_C_3 = 10
60_C_30 = 118264581564861424

TI-57

Machine code

Comment

Lbl 9
STO 2 
x⮂t 
STO 1
SBR 0
STO 3
RCL 1
- 
RCL 2
=
SBR 0
INV Prd 3
RCL 2
SBR 0
Inv Prd 3
RCL 3
R/S
RST
Lbl 0
C.t
x=t
1
STO 0
Lbl 1
RCL 0
×
Dsz
GTO 1
1
=
INV SBR

program binomial(x,t) // x is the display register
r2 = x
swap x and t
r1 = x
x = factorial(x)
r3 = x
 
 
 
x = r1 - r2
x = factorial(x)
r3 /= x
x = r2
x = factorial(x)
r3 /= x
x = r3
end program
reset pointer
program factorial(x) 
t=0
if x=0 then
  x=1
r0 = x
loop
  
  multiply r0 by what will be in the next loop
  decrement r0 and exit loop if r0 = 0
end loop
complete the multiplication sequence
return x!
end sub

5 x⮂t 3 GTO 9 R/S

Output:

10.

TI-83 BASIC

Builtin operator nCr gives the number of combinations.

10 nCr 4

Output:

TI-89 BASIC

Builtin function.

nCr(n,k)

TXR

nCk is a built-in function, along with the one for permutations, nPk:

$ txr -p '(n-choose-k 20 15)'
15504

$ txr -p '(n-perm-k 20 15)'
20274183401472000

UNIX Shell

#!/bin/sh                                                                                                                                             
n=5;
k=3;
calculate_factorial(){
partial_factorial=1;
for (( i=1; i<="$1"; i++ ))
do
    factorial=$(expr $i \* $partial_factorial)
    partial_factorial=$factorial

done
echo $factorial
}

n_factorial=$(calculate_factorial $n)
k_factorial=$(calculate_factorial $k)
n_minus_k_factorial=$(calculate_factorial `expr $n - $k`)
binomial_coefficient=$(expr $n_factorial \/ $k_factorial \* 1 \/ $n_minus_k_factorial )

echo "Binomial Coefficient ($n,$k) = $binomial_coefficient"

Ursala

A function for computing binomial coefficients (choose) is included in a standard library, but if it weren't, it could be defined in one of the following ways, starting from the most idiomatic.

#import nat

choose = ~&ar^?\1! quotient^\~&ar product^/~&al ^|R/~& predecessor~~

The standard library functions quotient, product and predecessor pertain to natural numbers in the obvious way.

choose is defined using the recursive conditional combinator (^?) as a function taking a pair of numbers, with the predicate ~&ar testing whether the number on the right side of the pair is non-zero.
If the predicate does not hold (implying the right side is zero), then a constant value of 1 is returned.
If the predicate holds, the function given by the rest of the expression executes as follows.
First the predecessor of both sides (~~) of the argument is taken.
Then a recursive call (^|R) is made to the whole function (~&) with the pair of predecessors passed to it as an argument.
The result returned by the recursive call is multiplied (product) by the left side of the original argument (~&al).
The product of these values is then divided (quotient) by the right side (~&ar) of the original argument and returned as the result.

Here is a less efficient implementation more closely following the formula above.

choose = quotient^/factorial@l product+ factorial^~/difference ~&r

choose is defined as the quotient of the results of a pair (^) of functions.
The left function contributing to the quotient is the factorial of the left side (@l) of the argument, which is assumed to be a pair of natural numbers. The factorial function is provided in a standard library.
The right function contributing to the quotient is the function given by the rest of the expression, which applies to the whole pair as follows.
It begins by forming a pair of numbers from the argument, the left being their difference obtained by subtraction, and the right being the a copy of the right (~&r) side of the argument.
The factorial function is applied separately to both results (^~).
A composition (+) of this function with the product function effects the multiplication of the two factorials, to complete the other input to the quotient.

Here is an equivalent implementation using pattern matching, dummy variables, and only the apply-to-both (~~) operator.

choose("n","k") = quotient(factorial "n",product factorial~~ (difference("n","k"),"k"))

test program:

#cast %nL

main = choose* <(5,3),(60,30)>

Output:

<10,118264581564861424>

VBScript

Function binomial(n,k)
	binomial = factorial(n)/(factorial(n-k)*factorial(k))
End Function

Function factorial(n)
	If n = 0 Then
		factorial = 1
	Else
		For i = n To 1 Step -1
			If i = n Then
				factorial = n
			Else
				factorial = factorial * i
			End If
		Next
	End If
End Function

'calling the function
WScript.StdOut.Write "the binomial coefficient of 5 and 3 = " & binomial(5,3)
WScript.StdOut.WriteLine

Output:

the binomial coefficient of 5 and 3 = 10

Wren

Library: Wren-fmt

Library: Wren-math

import "./fmt" for Fmt
import "./math" for Int

var binomial = Fn.new { |n, k|
    if (n < 0 || k < 0) Fiber.abort("Arguments must be non-negative integers")
    if (n < k) Fiber.abort("The second argument cannot be more than the first.")
    if (n == k) return 1
    var prod = 1
    var i = n - k + 1
    while (i <= n) {
        prod = prod * i
        i = i + 1
    }
    return prod / Int.factorial(k)
}

var limit = 14
System.write("n/k |")
for (k in 0..limit) System.write(Fmt.d(5, k))
System.print()
System.write("----+" + "-----" * (limit + 1))
System.print()
for (n in 0..limit) {
    System.write("%(Fmt.d(3, n)) |")
    for (k in 0..n) System.write(Fmt.d(5, binomial.call(n, k)))
    System.print()
}

Output:

n/k |    0    1    2    3    4    5    6    7    8    9   10   11   12   13   14
----+---------------------------------------------------------------------------
  0 |    1
  1 |    1    1
  2 |    1    2    1
  3 |    1    3    3    1
  4 |    1    4    6    4    1
  5 |    1    5   10   10    5    1
  6 |    1    6   15   20   15    6    1
  7 |    1    7   21   35   35   21    7    1
  8 |    1    8   28   56   70   56   28    8    1
  9 |    1    9   36   84  126  126   84   36    9    1
 10 |    1   10   45  120  210  252  210  120   45   10    1
 11 |    1   11   55  165  330  462  462  330  165   55   11    1
 12 |    1   12   66  220  495  792  924  792  495  220   66   12    1
 13 |    1   13   78  286  715 1287 1716 1716 1287  715  286   78   13    1
 14 |    1   14   91  364 1001 2002 3003 3432 3003 2002 1001  364   91   14    1

XPL0

code ChOut=8, CrLf=9, IntOut=11;

func Binomial(N, K);
int  N, K;
int  M, B, I;
[M:= K;
if K>N/2 the M:= N-K;
B:=1;
for I:= 1 to M do
    B:= B*(N-M+I)/I;
return B;
];

int N, K;
[for N:= 0 to 9 do
    [for K:= 0 to 9 do
        [if N>=K then IntOut(0, Binomial(N,K));
        ChOut(0, 9\tab\);
        ];
    CrLf(0);
    ];
] \Mr. Pascal's triangle!

Output:

1                                                                               
1       1                                                                       
1       2       1                                                               
1       3       3       1                                                       
1       4       6       4       1                                               
1       5       10      10      5       1                                       
1       6       15      20      15      6       1                               
1       7       21      35      35      21      7       1                       
1       8       28      56      70      56      28      8       1               
1       9       36      84      126     126     84      36      9       1

Zig

A reasonable implementation for a fixed word size is to precompute all possible values of nCk, since even on a 64 bit machine there's only 67 rows of Pascal's triangle to consider, or ~18K of data.

In Zig it's possible to compute all values of nCk at compile time, so that at runtime it's only necessary to do a table lookup. Zig also supports nullable values, so nCk can return a null value if the programmer requests a value that's out of range. Finally, since this code uses addition to compute the table, all entries that can fit in 64 bits can be computed, in contrast to some other code examples that may overflow before the maximum representable value (67 choose 33) is reached. For example, the largest value the BCPL version can compute is 61 choose 31.

const std = @import("std");

pub fn binomial(n: u32) ?[]const u64 {
    if (n >= rmax)
        return null
    else {
        const k = n * (n + 1) / 2;
        return pascal[k .. k + n + 1];
    }
}

pub fn nCk(n: u32, k: u32) ?u64 {
    if (n >= rmax)
        return null
    else if (k > n)
        return 0
    else {
        const j = n * (n + 1) / 2;
        return pascal[j + k];
    }
}        

const rmax = 68;

const pascal = build: {
    @setEvalBranchQuota(100_000);
    var coefficients: [(rmax * (rmax + 1)) / 2]u64 = undefined;
    coefficients[0] = 1;
    var j: u32 = 0;
    var k: u32 = 1;
    var n: u32 = 1;
    while (n < rmax) : (n += 1) {
        var prev = coefficients[j .. j + n];
        var next = coefficients[k .. k + n + 1];
        next[0] = 1;
        var i: u32 = 1;
        while (i < n) : (i += 1)
            next[i] = prev[i] + prev[i - 1];
        next[i] = 1;
        j = k;
        k += n + 1;
    }
    break :build coefficients;
};

test "n choose k" {
    const expect = std.testing.expect;
    try expect(nCk(10, 5).? == 252);
    try expect(nCk(10, 11).? == 0);
    try expect(nCk(10, 10).? == 1);
    try expect(nCk(67, 33).? == 14226520737620288370); 
    try expect(nCk(68, 34) == null);
}

Rather than write driver code, it's possible to run the unit test for this module.

Output:

$ zig test binomial.zig 
All 1 tests passed.

zkl

Using 64 bit ints:

fcn binomial(n,k){ (1).reduce(k,fcn(p,i,n){ p*(n-i+1)/i },1,n) }

Output:

zkl: binomial(5,3)
10
zkl: binomial(60,30)
118264581564861424

ZX Spectrum Basic

Translation of: BBC_BASIC

10 LET n=33: LET k=17: PRINT "Binomial ";n;",";k;" = ";
20 LET r=1: LET d=n-k
30 IF d>k THEN LET k=d: LET d=n-k
40 IF n<=k THEN GO TO 90
50 LET r=r*n
60 LET n=n-1
70 IF (d>1) AND (FN m(r,d)=0) THEN LET r=r/d: LET d=d-1: GO TO 70
80 GO TO 40
90 PRINT r
100 DEF FN m(a,b)=a-INT (a/b)*b