How to avoid redundant steps in calculation of number of permutations? [duplicate]

Question

How can I make the following script more intelligent, like a human would do it?

(avoiding the redundant calculations of the full factorials)

def combinations_without_repetition(n, r):
    return (factorial(n) / (factorial(r) * (factorial(n - r))))
combinations_without_repetition(10, 5)

(I get errors: Too large values.)

CDJB · Accepted Answer · 2019-11-20 15:31:53Z

1

You can use scipy.special.comb for this. Switching exact to True gives the exact answer as an integer at the cost of speed.

>>> from scipy.special import comb
>>> comb(10, 5, exact=False, repetition=False)
252.0
>>> comb(10, 5, exact=True, repetition=False)
252

answered Nov 20, 2019 at 15:31

CDJB

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