- scipy.special.comb(N, k, *, exact=False, repetition=False, legacy=<object object>)#
The number of combinations of N things taken k at a time.
This is often expressed as “N choose k”.
- 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).
- valint, float, ndarray
The total number of combinations.
Binomial coefficient considered as a function of two real variables.
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.
>>> 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