mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-21 02:30:15 +00:00
4700297b3e
* Enable ruff RUF002 rule * Fix --------- Co-authored-by: Christian Clauss <cclauss@me.com>
45 lines
959 B
Python
45 lines
959 B
Python
"""
|
|
Combinatoric selections
|
|
Problem 53
|
|
|
|
There are exactly ten ways of selecting three from five, 12345:
|
|
|
|
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
|
|
|
|
In combinatorics, we use the notation, 5C3 = 10.
|
|
|
|
In general,
|
|
|
|
nCr = n!/(r!(n-r)!),where r ≤ n, n! = nx(n-1)x...x3x2x1, and 0! = 1.
|
|
It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
|
|
|
|
How many, not necessarily distinct, values of nCr, for 1 ≤ n ≤ 100, are greater
|
|
than one-million?
|
|
"""
|
|
|
|
from math import factorial
|
|
|
|
|
|
def combinations(n, r):
|
|
return factorial(n) / (factorial(r) * factorial(n - r))
|
|
|
|
|
|
def solution():
|
|
"""Returns the number of values of nCr, for 1 ≤ n ≤ 100, are greater than
|
|
one-million
|
|
|
|
>>> solution()
|
|
4075
|
|
"""
|
|
total = 0
|
|
|
|
for i in range(1, 101):
|
|
for j in range(1, i + 1):
|
|
if combinations(i, j) > 1e6:
|
|
total += 1
|
|
return total
|
|
|
|
|
|
if __name__ == "__main__":
|
|
print(solution())
|