# Is there a math nCr function in python?

## Is there a math nCr function in python?

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
```

#### Is there a math nCr function in python?

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