In place of calculating the factorial several times we can run a loop k times to calculate the combination (#10051)

* In place of calculating the factorial several times we can run a loop k times to calculate the combination 

for example:
5 C 3 = 5! / (3! * (5-3)! )
= (5*4*3*2*1)/[(3*2*1)*(2*1)]
=(5*4*3)/(3*2*1)
so running a loop k times will reduce the time complexity to O(k)

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Update maths/combinations.py

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Tianyi Zheng <tianyizheng02@gmail.com>

Author: 3 people

maths/combinations.py

@@ -1,7 +1,6 @@
 """
 https://en.wikipedia.org/wiki/Combination
 """
-from math import factorial
 
 
 def combinations(n: int, k: int) -> int:
@@ -35,7 +34,11 @@ def combinations(n: int, k: int) -> int:
     # to calculate a factorial of a negative number, which is not possible
     if n < k or k < 0:
         raise ValueError("Please enter positive integers for n and k where n >= k")
-    return factorial(n) // (factorial(k) * factorial(n - k))
+    res = 1
+    for i in range(k):
+        res *= n - i
+        res //= i + 1
+    return res
 
 
 if __name__ == "__main__":