# Evaluate binomial coefficients

(Redirected from Binomial coefficients)
Evaluate binomial coefficients
You are encouraged to solve this task according to the task description, using any language you may know.

This programming task, is to calculate ANY binomial coefficient.

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

This formula is recommended:

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

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 ${\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}$ ${\displaystyle n^{k}}$

## 11l

Translation of: Python
F binomial_coeff(n, k)
V result = 1
L(i) 1..k
result = result * (n - i + 1) / i
R result

print(binomial_coeff(5, 3))
Output:
10


## 360 Assembly

Translation of: ABAP

Very compact version.

*        Evaluate binomial coefficients - 29/09/2015
BINOMIAL CSECT
USING  BINOMIAL,R15       set base register
SR     R4,R4              clear for mult and div
LA     R5,1               r=1
LA     R7,1               i=1
L      R8,N               m=n
LOOP     LR     R4,R7              do while i<=k
C      R4,K               i<=k
BH     LOOPEND            if not then exit while
MR     R4,R8              r*m
DR     R4,R7              r=r*m/i
LA     R7,1(R7)           i=i+1
BCTR   R8,0               m=m-1
B      LOOP               loop while
LOOPEND  XDECO  R5,PG              edit r
XPRNT  PG,12              print r
XR     R15,R15            set return code
N        DC     F'10'              <== input value
K        DC     F'4'               <== input value
PG       DS     CL12               buffer
YREGS
END    BINOMIAL
Output:
         210


## 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)))


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}

## Arturo

factorial: function [n]-> product 1..n
binomial:  function [x,y]-> (factorial x) / (factorial y) * factorial x-y

print binomial 5 3

Output:
10

## AutoHotkey

MsgBox, % Round(BinomialCoefficient(5, 3))

;---------------------------------------------------------------------------
BinomialCoefficient(n, k) {
;---------------------------------------------------------------------------
r := 1
Loop, % k < n - k ? k : n - k {
r *= n - A_Index + 1
r /= A_Index
}
Return, r
}


Message box shows:

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.

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);
}

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:
10

## 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:
10

### 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
do i=0,j
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 ${\displaystyle \prod _{i=0}^{k-1}{\frac {n-i}{i+1}}}$:

;5 3 # Set up demo input

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 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: 10 ## 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: 10 ## 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;

.[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: 10  ## 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: 10 ## 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: 10 ## 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: 10 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: 30101 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: 10  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:  10  ## 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: 10 2 1 45 45  ## 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: 10 ### 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: 10 ## 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: 10 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 1st REXX version. It is (around average) about ten times faster than the 1st 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: 10  ## 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: 10 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${\displaystyle ^{5}}$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: 10 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: 10 ## 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: 210  ## 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