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

### Describe your change:

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)

Author: SubhranShu2332

maths/combinations.py

@@ -1,9 +1,6 @@
 """
 https://en.wikipedia.org/wiki/Combination
 """
-from math import factorial
-
-
 def combinations(n: int, k: int) -> int:
     """
     Returns the number of different combinations of k length which can
@@ -35,8 +32,11 @@ 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 = res * (n - i)
+        res = res // (i + 1)
+    return res
 
 if __name__ == "__main__":
     print(