Still needing Binomial Code Help.....

[confusedndazed]

Hello people, this is my second attempt to get help for the same code. First post is unsolved. I apologize for any confusion in my posts but I'm confused myself so it's somewhat hard for me to properly explain that which I don't even understand....:$....

I have been given the following information:

/* pseudocode for Binomial
Coefficients */
int binomial(int n, int k)
{
If (n < k) Then return (0)
Else
{
Set denominator = 1*2*...*k
Set numerator = (n-k+1)*(n-k+2)*...*(n-1)*n
return (numerator / denominator)
} // else
End if
}

Here is my code thus far, sorry if it's really bad:

#include <iostream>

using namespace std ;

int binomial(int n, int k) ; // function prototype


int main ()
{
    
    int n, k ; // parameters for the binomial number
    int result ;
    
    cout << endl ;
    
    // read in n & k
    
    cout << "Enter n (positive integer) : " ;
    cin >> n ;
    
    cout << "Enter k (positive integer) : " ;
    cin >> k ;
    
    result = binomial(n,k); 
    
    cout << "Binomial number " << n << "C" << k
         << " = " << result << endl ;
         
    return (0) ;
}

int binomial(int n, int k)

{
    int numerator, denominator ;
    int i ; // needed to compute numerator & denominator
    
    if (n < k) Then
{
    return (0) ; 
}
  else
  {
      denominator = 1*2*...*k ; 
      
      for ( i = k : <= n ; i=i+1)
     denominator = * ;
     numerator = (n-k+1)*(n-k+2)*...*(n-1)*n

     
     return (numerator / denominator) ; 
}

I am very new and lost so please bear with me in your replies, as I still may not "get it".....Thanks!!

[mrnutty]

>>denominator = 1*2*...*k

I hope you know thats not valid. 1 * 2 * 3 ... k is called the factorial of K

So if K was 5, the the denominator will be 1 * 2 * 3 * 4 * 5.

So make a function : unsigned long factorial(unsigned long num);
Where it returns the factorial of the parameters passed. So if num
was 5, then the call would be : denominator = factorial(5); // = 120

Fix that then post back, because there are other problems.

[Agni]

By binomial coefficient you mean finding the combinatorial number, is that right ? So you should be using the formula n!/(n-k)!*k!

for this the pseudocode should be

numerator=1*2*3..*n
fact1 = 1*2*3...*n-k
fact2 = 1*2*3....*k
denominator = fact1 * fact2
binomial coefficient = numerator/denominator

can you code that?

[lotrsimp12345]

you probably need a factorial function to find 1*2...*k.

it should probably be recursive

[confusedndazed]

>>denominator = 1*2*...*k

I hope you know thats not valid. 1 * 2 * 3 ... k is called the factorial of K

So if K was 5, the the denominator will be 1 * 2 * 3 * 4 * 5.

So make a function : unsigned long factorial(unsigned long num);
Where it returns the factorial of the parameters passed. So if num
was 5, then the call would be : denominator = factorial(5); // = 120

Fix that then post back, because there are other problems.

Thanks but how do I find out what number k is? You used 5 as an example right? I'm really confused about this....

[lotrsimp12345]

you pass in k to your function i believe.