scipy.special.comb#

scipy.special.comb(N, k, *, exact=False, repetition=False, legacy=<object object>)[source]#

The number of combinations of N things taken k at a time.

This is often expressed as “N choose k”.

Parameters:

Nint, ndarray: Number of things.
kint, ndarray: Number of elements taken.
exactbool, optional: For integers, if exact is False, then floating point precision is used, otherwise the result is computed exactly. For non-integers, if exact is True, is disregarded.
repetitionbool, optional: If repetition is True, then the number of combinations with repetition is computed.
legacybool, optional: If legacy is True and exact is True, then non-integral arguments are cast to ints; if legacy is False, the result for non-integral arguments is unaffected by the value of exact.

Deprecated since version 1.9.0: Using legacy is deprecated and will removed by Scipy 1.14.0. If you want to keep the legacy behaviour, cast your inputs directly, e.g. comb(int(your_N), int(your_k), exact=True).

Returns:

valint, float, ndarray: The total number of combinations.

See also

binom: Binomial coefficient considered as a function of two real variables.

Notes

Array arguments accepted only for exact=False case.
If N < 0, or k < 0, then 0 is returned.
If k > N and repetition=False, then 0 is returned.

Examples

>>> import numpy as np
>>> from scipy.special import comb
>>> k = np.array([3, 4])
>>> n = np.array([10, 10])
>>> comb(n, k, exact=False)
array([ 120.,  210.])
>>> comb(10, 3, exact=True)
120
>>> comb(10, 3, exact=True, repetition=True)
220