Is there a math nCr function in Python?

Question

Is there a built-in nCr (n choose r) function included in the Python math library like the one shown below?

I understand that the computation can be programmed, but I thought I’d check to see if it’s built-in before I do.

||

Answer 1

Do you want iteration? Use itertools.combinations. Common usage:

>>> import itertools
>>> itertools.combinations('abcd', 2)
<itertools.combinations object at 0x104e9f010>
>>> list(itertools.combinations('abcd', 2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd', 2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']

If you just need to compute the formula, math.factorial can be used, but is not fast for large combinations, but see math.comb below for an optimized calculation available in Python 3.8+:

import math

def ncr(n, r):
    f = math.factorial
    return f(n) // f(r) // f(n-r)

print(ncr(4, 2))  # Output: 6

As of Python 3.8, math.comb can be used and is much faster:

>>> import math
>>> math.comb(4,2)
6

Answered By: Mark Tolonen

Answer 2

On Python 3.8+, use math.comb:

>>> from math import comb
>>> comb(10, 3)
120

For older versions of Python, you can use the following program:

import operator as op
from functools import reduce

def ncr(n, r):
    r = min(r, n-r)
    numer = reduce(op.mul, range(n, n-r, -1), 1)
    denom = reduce(op.mul, range(1, r+1), 1)
    return numer // denom  # or / in Python 2

Answered By: dheerosaur

Is there a math nCr function in Python?

Question:

Answers: