## Problem:

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! = n[×](n[−]1)[×]...[×]3[×]2[×]1, 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?

6857

## Code:The solution may include methods that will be found here: Library.java .

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 3 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p003 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p003().run());	}			/* 	 * Algorithm: Divide out all the smallest prime factors except the last one.	 * For example, 1596 = 2 * 2 * 3 * 7 * 19. The algorithm ensures that the smallest factors will be found first.	 * After dividing out the smallest factors, the last factor to be found will be equal to the quotient, so it must be the largest prime factor.	 */	public String run() {		long n = 600851475143L;		while (true) {			long p = smallestFactor(n);			if (p < n)				n /= p;			else				return Long.toString(n);		}	}			private static long smallestFactor(long n) {		for (long i = 2, end = Library.sqrt(n); i <= end; i++) {			if (n % i == 0)				return i;		}		return n;  // Prime	}	}