Solution to Project Euler Problem 53: Combinatoric selections - 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 C 3) = 10.
In general. (n C r) = n!/r!(n-r)!, where r ≤ n, n! = n x (n - 1) x ... 3 x 2 x 1, and 0! = 1.

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

import java.math.BigInteger;

public class SikademyEulerSolution {
	
	public String run() {
		BigInteger MILLION = BigInteger.TEN.pow(6);
		int count = 0;
		for (int n = 1; n <= 100; n++) {
			for (int r = 0; r <= n; r++) {
				if (binomial(n, r).compareTo(MILLION) > 0)
					count++;
			}
		}
		return Integer.toString(count);
	}

        public static BigInteger binomial(int n, int k) {
		if (k < 0 || k > n)
			throw new IllegalArgumentException();
		BigInteger product = BigInteger.ONE;
		for (int i = 0; i < k; i++)
			product = product.multiply(BigInteger.valueOf(n - i));
		return product.divide(factorial(k));
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

import math

def compute():
	ans = sum(1
		for n in range(1, 101)
		for k in range(0, n + 1)
		if binomial(n, k) > 1000000)
	return str(ans)


def binomial(n, k):
	assert 0 <= k <= n
	return math.factorial(n) // (math.factorial(k) * math.factorial(n - k))

if __name__ == "__main__":
	print(compute())