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

If I understand the notation in the question, you don't just want nCp, you actually want all of nC1, nC2, ... nC(n-1). If this is correct, we can leverage the following relationship to make this fairly trivial:

• for all k>0: nCk = prod_{from i=1..k}( (n-i+1)/i )
• i.e. for all k>0: nCk = nC(k-1) * (n-k+1) / k

Here's a python snippet implementing this approach:

def binomial_coef_seq(n, k):
    """Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""
    b = [1]
    for i in range(1,k+1):
        b.append(b[-1] * (n-i+1)/i)
    return b

If you need all coefficients up to some k > ceiling(n/2), you can use symmetry to reduce the number of operations you need to perform by stopping at the coefficient for ceiling(n/2) and then just backfilling as far as you need.

import numpy as np

def binomial_coef_seq2(n, k):
    """Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""

    k2 = int(np.ceiling(n/2))

    use_symmetry =  k > k2
    if use_symmetry:
        k = k2

    b = [1]
    for i in range(1, k+1):
        b.append(b[-1] * (n-i+1)/i)

    if use_symmetry:
        v = k2 - (n-k)
        b2 = b[-v:]
        b.extend(b2)
    return b