A fast way to calculate binomial coefficients in python (Andrew Dalke)

binomial.py

def binomial(n, k):
    """
    A fast way to calculate binomial coefficients by Andrew Dalke.
    See http://stackoverflow.com/questions/3025162/statistics-combinations-in-python
    """
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in xrange(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def binom(n,k): # better version - we don't need two products!
if not 0<=k<=n: return 0
b=1
for t in range(min(k,n-k)):
b*=n; b/=t+1; n-=1
return b

Formatted:

def binomial(n, k):
    if not 0 <= k <= n:
        return 0
    b = 1
    for t in range(min(k, n-k)):
        b *= n
        b /= t+1
        n -= 1
    return int(b)

So for example when you call binomial(5, 2) it returns 10.

You can use b //= t+1 to avoid final cast.

The intention was that this should use only integer arithmetic (my version was converted from C code which used /=). So yes, this is better:

def binomial(n, k):
    if not 0 <= k <= n:
        return 0
    b = 1
    for t in range(min(k, n-k)):
        b *= n
        b //= t+1
        n -= 1
    return b

def C(n, k): if k == 0 or k == n: return 1 else: return C(n-1, k) + C(n-1, k-1)

It is much more efficient NOT to use recursion.

Try this....a short way by importing comb from math library

from math import comb
def coefficient(n,r):
return comb(n,r)
print(coefficient(5,2))

Change coefficient values to your desire