# Is there a math nCr function in python? [duplicate]

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I’m looking to see if built in with the math library in python is the nCr (n Choose r) function: I understand that this can be programmed but I thought that I’d check to see if it’s already built in before I do.

The following program calculates `nCr` in an efficient manner (compared to calculating factorials etc.)

``````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
``````

As of Python 3.8, binomial coefficients are available in the standard library as `math.comb`:

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

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

``````>>> import itertools
>>> itertools.combinations('abcd',2)
<itertools.combinations object at 0x01348F30>
>>> 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, use math.factorial:

``````import math

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

if __name__ == '__main__':
print nCr(4,2)
``````

In Python 3, use the integer division `//` instead of `/` to avoid overflows:

`return f(n) // f(r) // f(n-r)`

### Output

``````6
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