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Programming-Idioms

  • Pascal
  • Scheme
  • Java

Idiom #67 Binomial coefficient "n choose k"

Calculate binom(n, k) = n! / (k! * (n-k)!). Use an integer type able to handle huge numbers.

import java.math.BigInteger;
static BigInteger binom(int N, int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}
uses ncalc;
var
  N, K, Res: TValue;
begin
  Res := BinomN(N, K);
end.

Pascal does not implement a native "big integer" type.
There are several 3rd party libraries that do so. One of them is ncalc, which defines TValue which can hold up to 2 gigabyte decimals.
long long binom(long long n, long long k){
  long long outp = 1;
  for(long long i = n - k + 1; i <= n; ++i) outp *= i;
  for(long long i = 2; i <= k; ++i) outp /= i;
  return outp;
}

c does not have a BigInt type, as a substitute long long is used.
To increase the range as much as possible, the formula is slightly different than the one presented in the idiom, but it gives the same results.
New implementation...
< > ⌨
programming-idioms.org 
static BigInteger binom(int N, int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}
var
  N, K, Res: TValue;
begin
  Res := BinomN(N, K);
end.
long long binom(long long n, long long k){
  long long outp = 1;
  for(long long i = n - k + 1; i <= n; ++i) outp *= i;
  for(long long i = 2; i <= k; ++i) outp /= i;
  return outp;
}
(defn binom [n k]
  (let [fact #(apply * (range 1 (inc %)))]
    (/ (fact n)
       (* (fact k) (fact (- n k))))))
public BigInteger binom(int n, int k)
{
	return factorial(n)/(factorial(k) * factorial(n-k));
}

public BigInteger factorial(int x)
{
	BigInteger result = 1;
	for(int i=1;i<=x;i++)
	{
	result = result * i;
	}
	return result;
}

BigInt binom(uint n, uint k)
{
   assert(n >= k);
   BigInt r = 1;
   for(uint i = 0; i <= k; ++i)
   {
      r *= n-i;
      r /= i+1;
   }
   return r;
}
int binom(int n, int k) {
  int result = 1;
  for (int i = 0; i < k; i++) {
    result = result * (n - i) ~/ (i + 1);
  }
  return result;
}
integer, parameter :: i8 = selected_int_kind(18)
integer, parameter :: dp = selected_real_kind(15)
n = 100
k = 5
print *,nint(exp(log_gamma(n+1.0_dp)-log_gamma(n-k+1.0_dp)-log_gamma(k+1.0_dp)),kind=i8)
z := new(big.Int)
z.Binomial(n, k)
binom n k = product [1+n-k..n] `div` product [1..k]

const fac = x => x ? x * fac (x - 1) : x + 1
const binom = (n, k) => fac (n) / fac (k) / fac (n - k >= 0 ? n - k : NaN)
gmp_binomial($n, $k);
sub binom {
  my ($n, $k) = @_;
  if ($k > $n - $k) { $k = $n - $k }
  my $r = 1;
  for ( my $i = $n/$n ; $i <= $k;) {
    $r *= $n-- / $i++
  }
  return $r
}
sub binom {
   my ($n, $k) = @_;
   my $fact = sub {
      my $n = shift;
      return $n<2 ? 1 : $n * $fact->($n-1);
   };
   return $fact->($n) / ($fact->($k) * ($fact->($n-$k)));
}
def binom(n, k):
    return math.factorial(n) // math.factorial(k) // math.factorial(n - k)
def binom(n, k):
    return math.comb(n, k)
def binom(n,k)
  (1+n-k..n).inject(:*)/(1..k).inject(:*)
end
fn binom(n: u64, k: u64) -> BigInt {
    let mut res = BigInt::one();
    for i in 0..k {
        res = (res * (n - i).to_bigint().unwrap()) /
              (i + 1).to_bigint().unwrap();
    }
    res
}