Evaluate 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 ${\binom {5}{3}}$, which is 10.

This formula is recommended:

${\binom {n}{k}}={\frac {n!}{(n-k)!k!}}$

C++

<lang cpp>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 EvaluateBinomialCoefficient(double nValue, double nValue2)

  {
      double result;

      result = (Factorial(nValue))/(Factorial(nValue2)*Factorial((nValue - nValue2)));
      nValue2 = result;
      return nValue2;
  }</lang>

Implementation: <lang cpp>int main() {

   cout<<"The Binomial Coefficient of 5, and 3, is equal to: "<< EvaluateBinomialCoefficient(5,3);
   cin.get();

}</lang>

Output:

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

Java

<lang java>public class Binom {

   public static double fact(double a) {
       double retVal = 1;
       for (double i = a; i > 1; i--) {
           retVal *= i;
       }
       return retVal;
   }

   public static double binomialCoeff(double n, double k) {
       return (fact(n)) / (fact(k) * fact((n - k)));
   }

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

}</lang> Output:

10.0

Translation of: Python

<lang java>public class Binom {

   public static double binomCoeff(double n, double k) {
       double result = 1;
       for (int i = 1; i < k + 1; i++) {
           result *= (n - i + 1) / i;
       }
       return result;
   }

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

} </lang> Output:

10.0

Python

an efficient implementation: <lang python>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))</lang>

Output:

10

Tcl

This uses exact arbitrary precision integer arithmetic. <lang tcl>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}]

}</lang> Demonstrating: <lang tcl>puts "60_C_30 = [binom 60 30]"</lang> Output:

60_C_30 = 118264581564861424