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Number of ways to go from one point to another in a grid

  • Difficulty Level : Easy
  • Last Updated : 16 Apr, 2021

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.
 

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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
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// function that will
// calculate the factorial
function factorial(N)
{
    var result = 1;
    while (N > 0) {
        result = result * N;
        N--;
    }
    return result;
}
 
function countWays(N)
{
    var total = factorial(N + N);
    var total1 = factorial(N);
    return (total / total1) / total1;
}
 
// Driver code
var N = 5;
document.write( "Ways = " + countWays(N));
 
// This code is contributed by rutvik_56.
</script>
Output: 
Ways = 252

 




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