"""
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, (5 3) = 10 .

In general, (n r) = n! / (r! * (n - r)!) , where r ≤ n , n!=n×(n-1)×...×3×2×1 , and 0!=1 .

It is not until n=23 , that a value exceeds one-million: (23 10) = 1144066 .

How many, not necessarily distinct, values of (n r) for 1 ≤ n ≤ 100 , are greater than one-million?
"""

import time
from functools import lru_cache


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.time()
        result = func(*args, **kwargs)
        end_time = time.time()
        elapsed_time = end_time - start_time
        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
        return result

    return wrapper


def comfun(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r))


@lru_cache(maxsize=None)
def factorial(n: int, s: int = 1):
    if n < s:
        return 1
    elif n == s:
        if s < 1:
            return 1
        return s
    else:
        return n * factorial(n - 1)


@timer
def main():
    num = 0
    for i in range(1, 101):
        for j in range(1, i + 1):
            if comfun(i, j) > 1000000:
                num += 1
    print(num)


if __name__ == "__main__":
    main()