#!/usr/bin/python

# 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?

from scipy.special import comb

from projecteuler import timing


@timing
def p053() -> None:
    LIMIT = 1000000

    count = 0

    # Use the scipy comb function to calculate the binomial values
    for i in range(23, 101):
        for j in range(1, i+1):
            if comb(i, j) > LIMIT:
                count = count + 1

    print('Project Euler, Problem 53')
    print(f'Answer: {count}')


if __name__ == '__main__':
    p053()