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Number of ways to go from one point to another in a grid
  • Difficulty Level : Easy
  • Last Updated : 08 May, 2019

Given the NxN grid of horizontal and vertical roads. The task is to find out the number of ways that the person can go from point A to point B using the shortest possible path.

Note: A and B point are fixed i.e A is at top left corner and B at bottom right corner as shown in the below image.

In the above image, the path shown in the red and light green colour are the two possible paths to reach from point A to point B.

Examples:



Input: N = 3
Output: Ways = 20

Input: N = 4
Output: Ways = 70

Formula:
Let the grid be N x N, number of ways can be written as.

How does above formula work?
Let consider the example of the 5×5 grid as shown above. In order to go from point A to point B in the 5×5 grid, We have to take 5 horizontal steps and 5 vertical steps. Each path will be an arrangement of 10 steps out of which 5 steps are identical of one kind and other 5 steps are identical of a second kind. Therefore

No. of ways = 10! / (5! * 5!) i.e 252 ways.

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// function that will
// calculate the factorial
long factorial(int N)
{
    int result = 1;
    while (N > 0) {
        result = result * N;
        N--;
    }
    return result;
}
  
long countWays(int N)
{
    long total = factorial(N + N);
    long total1 = factorial(N);
    return (total / total1) / total1;
}
  
// Driver code
int main()
{
    int N = 5;
    cout << "Ways = " << countWays(N);
    return 0;
}


Java




// Java implementation of above approach
class GfG {
  
    // function that will
    // calculate the factorial
    static long factorial(int N)
    {
        int result = 1;
        while (N > 0) {
            result = result * N;
            N--;
        }
        return result;
    }
  
    static long countWays(int N)
    {
        long total = factorial(N + N);
        long total1 = factorial(N);
        return (total / total1) / total1;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println("Ways = " + countWays(N));
    }
}


Python3




# Python3 implementation of above approach 
  
# function that will calculate the factorial 
def factorial(N) : 
      
    result = 1
      
    while (N > 0) :
          
        result = result * N; 
        N -= 1
      
    return result; 
  
def countWays(N) : 
  
    total = factorial(N + N); 
    total1 = factorial(N); 
      
    return (total // total1) // total1; 
  
# Driver code 
if __name__ == "__main__"
  
    N = 5
      
    print("Ways =", countWays(N)); 
  
# This code is contributed by Ryuga


C#




// C# implementation of above approach 
using System; 
class GfG 
  
    // function that will 
    // calculate the factorial 
    static long factorial(int N) 
    
        int result = 1; 
        while (N > 0) 
        
            result = result * N; 
            N--; 
        
        return result; 
    
  
    static long countWays(int N) 
    
        long total = factorial(N + N); 
        long total1 = factorial(N); 
        return (total / total1) / total1; 
    
  
    // Driver code 
    public static void Main(String []args) 
    
        int N = 5; 
        Console.WriteLine("Ways = " + countWays(N)); 
    
  
// This code is contributed by Arnab Kundu


PHP




<?php
// PHP implementation of above approach
  
// function that will
// calculate the factorial
function factorial($N)
{
    $result = 1;
    while ($N > 0) 
    {
        $result = $result * $N;
        $N--;
    }
    return $result;
}
  
function countWays($N)
{
    $total = factorial($N + $N);
    $total1 = factorial($N);
    return ($total / $total1) / $total1;
}
  
// Driver code
$N = 5;
echo "Ways = ", countWays($N);
      
// This code is contributed by ajit
?>


Output:

Ways = 252

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