⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

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

This formula is recommended: :: $\binom\left\{n\right\}\left\{k\right\} = \frac\left\{n!\right\}\left\{\left(n-k\right)!k!\right\} = \frac\left\{n\left(n-1\right)\left(n-2\right)\ldots\left(n-k+1\right)\right\}\left\{k\left(k-1\right)\left(k-2\right)\ldots 1\right\}$

• [[Combinations and permutations]]
• [[Pascal's triangle]] {{Template:Combinations and permutations}}

11l

{{trans|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))


{{out}}


10



360 Assembly

{{trans|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


{{out}}


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.


{{Out}}

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)))



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;



{{Out}}


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 === {{trans|C}} - note: This specimen retains the original [[Evaluate binomial coefficients#C|C]] coding style.

{{works with|ALGOL 68|Revision 1 - no extensions to language used}}

{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 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))
)


{{Out}}


+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.


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


{{Out}}

{10, 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)
}



{{out}}


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


{{Out}}


> 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.

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%



{{Out}}

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; }  {{out}} 10 131282408400 11923179284862717872  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(); }  {{Out}} The Binomial Coefficient of 5, and 3, is equal to: 10  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; } } }  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)}"  {{Out}}  > 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  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)); }  {{out}} ( 5 3) = 2 (100 2) = 50 (100 98) = 50 (100 50) = 1976664223067613962806675336  The above wouldn't work for me (100C50 correctly gives 100891344545564193334812497256). This next one is a 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; }  {{out}} 120 100891344545564193334812497256  dc [sx1q]sz[d0=zd1-lfx*]sf[skdlfxrlk-lfxlklfx*/]sb  Demonstration:  <tt>10</tt> Annotated version: dc [ 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));
end.


Elixir

{{trans|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)


{{out}}


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).



PROGRAM BINOMIAL



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)


{{Out}}

10


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


link math, factors

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


{{Out}}

choose(5,3)=10


{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/math.icn math provides binocoef] and [http://www.cs.arizona.edu/icon/library/src/procs/factors.icn 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


=={{header|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 <code>!</code> ([[http://www.jsoftware.com/help/dictionary/d410.htm Out of]]) evaluates binomial coefficients directly.

'''Example usage:'''

j
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);
}
}


{{Out}}

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));
}
}


{{Out}}

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));


{{Out}}

10


jq

= 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  {{out}} 5 C 3 = 10 100 C 2 = 4950 33 C 17 = 1166803110 Julia {{works with|Julia|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)  {{out}} 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 1.0.5-2 fun factorial(n: Int) = when { n < 0 -> throw IllegalArgumentException("negative numbers not allowed") else -> { var ans = 1L for (i in 2..n) ans *= i ans } } fun binomial(n: Int, k: Int) = when { n < 0 || k < 0 -> throw IllegalArgumentException("negative numbers not allowed") n == k -> 1L else -> { var ans = 1L for (i in n - k + 1..n) ans *= i ans / factorial(k) } } fun main(args: Array<String>) { for (n in 0..14) { for (k in 0..n) print("%4d ".format(binomial(n, k))) println() } }  {{out}}  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  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)  {{Out}} 10 5 118264581564861424  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  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  Maple convert(binomial(n,k),factorial); binomial(5,3);  {{Out}}  factorial(n) ----------------------------- factorial(k) factorial(n - k) 10  =={{header|Mathematica}} / {{header|Wolfram Language}}== (Local) In[1]:= Binomial[5,3] (Local) Out[1]= 10  =={{header|MATLAB}} / {{header|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|0.19.3}} ((dup 0 ==) 'succ (dup pred) '* linrec) :fact ('dup dip dup ((fact) () (- fact) (fact * div)) spread) :binomial 5 3 binomial puts!  {{out}}  10  =={{header|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. ## MINIL 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. Nim proc binomialCoeff(n, k: int): int = result = 1 for i in 1..k: result = result * (n-i+1) div i echo binomialCoeff(5, 3)  {{Out}} 10  Oberon {{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.  {{Out}} 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 / ] ;  {{out}}  >5 3 binomial . 10  Oz {{trans|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 [[Evaluate_binomial_coefficients#Delphi | Delphi]] Perl sub binomial { use bigint; my (r, n, k) = (1, @_); for (1 .. k) { r *= n--; r /= _ } r; } print binomial(5, 3);  {{out}} 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: {{libheader|ntheory}} use ntheory qw/binomial/; print length(binomial(100000,50000)), "\n";  {{out}} 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). Perl 6 For a start, you can get the length of the corresponding list of combinations: say combinations(5, 3).elems;  {{out}} 10  This method is efficient, as Perl 6 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:  { [*] (^n ... 0) Z/ 1 .. ^p } say 5 choose 3;  A possible optimization would use a symmetry property of the binomial coefficient:  { [*] (^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:  { ([*] (^n ... 0) Z/ 1 .. min(n - ^p, p)).Int }  And ''this'' is exactly what the count-only method does. Phix A naive version: function binom(integer n, k) return factorial(n)/(factorial(k)*factorial(n-k)) end function ?binom(5,3)  {{out}}  10  However errors will creep in should any result or interim value exceed 9,007,199,254,740,992 (on 32-bit), so (and using a different algorithm just for kicks): {{libheader|mpfr}} include builtins\mpfr.e function mpz_binom(integer n, k) mpz r = mpz_init(1) for i=1 to k do mpz_mul_si(r,r,n-i+1) if mpz_fdiv_q_ui(r, r, i)!=0 then ?9/0 end if -- r = ba_divide(ba_multiply(r,n-i+1),i) end for return mpz_get_str(r) end function ?mpz_binom(5,3) ?mpz_binom(100,50) ?mpz_binom(60,30) ?mpz_binom(1200,120)  {{out}}  "10" "100891344545564193334812497256" "118264581564861424" "1004576581793084916475353119318331966507299414258370667602185866686463289093457468590558508056798211449853806741873396451735444387513582540860551330127062642417424083600"  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; }  PicoLisp (de binomial (N K) (let f '((N) (if (=0 N) 1 (apply * (range 1 N))) ) (/ (f N) (* (f (- N K)) (f K)) ) ) )  {{Out}} : (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;  {{Out}}  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))  {{Out}} 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|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()  {{Out}} 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))))  {{Out}} 10 [1, 5, 10, 10, 5, 1]  R R's built-in choose() function evaluates binomial coefficients: choose(5,3)  {{Out}} [1] 10  Racket  #lang racket (require math) (binomial 10 5)  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  Ruby {{trans|Tcl}} {{works with|Ruby|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  {{Out}} 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)); }  {{Out}} 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) }  {{Out}} 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))) }  {{Out}} 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))  {{Out}} bico(5,3) = 10  Scheme {{Works with|Scheme|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)  {{Out}} 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)  {{Out}} 10  Seed7 The infix operator [http://seed7.sourceforge.net/libraries/integer.htm#%28in_integer%29!%28in_integer%29 !] 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;  {{Out}}  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 [http://seed7.sourceforge.net/libraries/bigint.htm bigint.s7i] contains a definition of the binomial coefficient operator [http://seed7.sourceforge.net/libraries/bigint.htm#%28in_bigInteger%29!%28in_var_bigInteger%29 !] for the type [http://seed7.sourceforge.net/manual/types.htm#bigInteger 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 [http://seed7.sourceforge.net/algorith/math.htm#binomial_coefficient]. 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)  Stata Use the [http://www.stata.com/help.cgi?comb comb] function. Notice the result is a missing value if k>n or k<0. . display comb(5,3) 10  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]"  {{Out}} 5_C_3 = 10 60_C_30 = 118264581564861424  =={{header|TI-83 BASIC}}== Builtin operator nCr gives the number of combinations.  {{out}} txt 210  =={{header|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)>  {{Out}} <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  {{Out}} the binomial coefficient of 5 and 3 = 10  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!


{{Out}}


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



zkl

Using 64 bit ints:

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


{{out}}


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



ZX Spectrum Basic

{{trans|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
`