Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20 x 20 grid?

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After doing a bit of research on this problem, it seems as though a simple *Combinations* formula will provide the answer. We know that any path will have 40 moves (20 right + 20 down), so for C(n,r) – n will equal 40 and r (the number of right moves) will equal 20 – C(40,20). Then we just plug the numbers into the formula – the ! means ‘factorial’. I posted 2 versions of the solution: 1 without recursion – uses a for loop, and one with recursion – the *factorial* function calls itself.

n!

C(n,r) = ———-

r!*(n-r)!

The number of how many good routes we have can be found by finding how

many combinations of 20 R’s we can have in our 40 moves, so we want to

calculate:

40!

C(40,20) = ———-

20!*(40-20)!

Helpful Links:

### One Possible Solution: Java – without recursion – (uses a for loop)

import java.math.BigInteger; public class Problem_15 { public static void main(String args[]){ int n = 40; // The total number of moves for any one path (right + down) int r = 20; // The total number of right moves for any one path System.out.println(factorial(n).divide(factorial(r).multiply(factorial(n - r)))); } public static BigInteger factorial( int n1 ) { BigInteger n = BigInteger.ONE; for (int i = 1; i <= n1; i ++) { n = n.multiply(BigInteger.valueOf(i)); } return n; } }

### One Possible Solution: Java – uses recursion (the factorial method calls itself)

import java.math.BigInteger; public class Problem_15a { public static void main(String args[]){ BigInteger n = BigInteger.valueOf(40); // The total number of moves for any one path (right + down) BigInteger r = BigInteger.valueOf(20); // The total number of right moves for any one path System.out.println(factorial(n).divide(factorial(r).multiply(factorial(n.subtract(r))))); } public static BigInteger factorial( BigInteger n ) { if ( n.compareTo(BigInteger.ZERO) <= 0 ) { // base case return BigInteger.ONE; } else { // general case return ( n.multiply(factorial ( n.subtract(BigInteger.ONE)) ) ); } } }