Fastest way to generate binomial coefficients

Question

I need to calculate combinations for a number.

What is the fastest way to calculate nCp where n>>p?

I need a fast way to generate binomial coefficients for a

Answer 1

I was looking for the same thing and couldn't find it, so wrote one myself that seems optimal for any Binomial Coeffcient for which the endresult fits into a Long.

// Calculate Binomial Coefficient 
// Jeroen B.P. Vuurens
public static long binomialCoefficient(int n, int k) {
    // take the lowest possible k to reduce computing using: n over k = n over (n-k)
    k = java.lang.Math.min( k, n - k );

    // holds the high number: fi. (1000 over 990) holds 991..1000
    long highnumber[] = new long[k];
    for (int i = 0; i < k; i++)
        highnumber[i] = n - i; // the high number first order is important
    // holds the dividers: fi. (1000 over 990) holds 2..10
    int dividers[] = new int[k - 1];
    for (int i = 0; i < k - 1; i++)
        dividers[i] = k - i;

    // for every dividers there is always exists a highnumber that can be divided by 
    // this, the number of highnumbers being a sequence that equals the number of 
    // dividers. Thus, the only trick needed is to divide in reverse order, so 
    // divide the highest divider first trying it on the highest highnumber first. 
    // That way you do not need to do any tricks with primes.
    for (int divider: dividers) {
       boolean eliminated = false;
       for (int i = 0; i < k; i++) {
          if (highnumber[i] % divider == 0) {
             highnumber[i] /= divider;
             eliminated = true;
             break;
          }
       }
       if(!eliminated) throw new Error(n+","+k+" divider="+divider);
    }


    // multiply remainder of highnumbers
    long result = 1;
    for (long high : highnumber)
       result *= high;
    return result;
}