# Counts paths from a point to reach Origin

You are standing on a point **(n, m)** and you want to go to origin **(0, 0)** by taking steps either **left or down** i.e. from each point you are allowed to move either in **(n-1, m)** or **(n, m-1)**. Find the number of paths from point to origin.

**Examples:**

Input : 3 6 Output : Number of Paths 84 Input : 3 0 Output : Number of Paths 1

As we are restricted to move down and left only we would run a recursive loop for each of the combinations of the

steps that can be taken.

// Recursive function to count number of paths countPaths(n, m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n==0 || m==0) return 1; // Else count sum of both ways return (countPaths(n-1, m) + countPaths(n, m-1)); }

Below is the implementation of the above steps.

## C++

// C++ program to count total number of // paths from a point to origin #include<bits/stdc++.h> using namespace std; // Recursive function to count number of paths int countPaths(int n, int m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n==0 || m==0) return 1; // Else count sum of both ways return (countPaths(n-1, m) + countPaths(n, m-1)); } // Driver Code int main() { int n = 3, m = 2; cout << " Number of Paths " << countPaths(n, m); return 0; }

## Java

// Java program to count total number of // paths from a point to origin import java.io.*; class GFG { // Recursive function to count number of paths static int countPaths(int n, int m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n == 0 || m == 0) return 1; // Else count sum of both ways return (countPaths(n - 1, m) + countPaths(n, m - 1)); } // Driver Code public static void main (String[] args) { int n = 3, m = 2; System.out.println (" Number of Paths " + countPaths(n, m)); } } // This code is contributed by vt_m

## Python3

# Python3 program to count # total number of # paths from a point to origin # Recursive function to # count number of paths def countPaths(n,m): # If we reach bottom # or top left, we are # have only one way to reach (0, 0) if (n==0 or m==0): return 1 # Else count sum of both ways return (countPaths(n-1, m) + countPaths(n, m-1)) # Driver Code n = 3 m = 2 print(" Number of Paths ", countPaths(n, m)) # This code is contributed # by Azkia Anam.

## C#

// C# program to count total number of // paths from a point to origin using System; public class GFG { // Recursive function to count number // of paths static int countPaths(int n, int m) { // If we reach bottom or top left, // we are have only one way to // reach (0, 0) if (n == 0 || m == 0) return 1; // Else count sum of both ways return (countPaths(n - 1, m) + countPaths(n, m - 1)); } // Driver Code public static void Main () { int n = 3, m = 2; Console.WriteLine (" Number of" + " Paths " + countPaths(n, m)); } } // This code is contributed by Sam007.

## PHP

<?php // PHP program to count total number // of paths from a point to origin // Recursive function to // count number of paths function countPaths($n, $m) { // If we reach bottom or // top left, we are // have only one way to // reach (0, 0) if ($n == 0 || $m == 0) return 1; // Else count sum of both ways return (countPaths($n - 1, $m) + countPaths($n, $m - 1)); } // Driver Code $n = 3; $m = 2; echo " Number of Paths " , countPaths($n, $m); // This code is contributed by aj_36 ?>

## Javascript

<script> // Javascript program to count total number of // paths from a point to origin // Recursive function to count number of paths function countPaths( n , m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n == 0 || m == 0) return 1; // Else count sum of both ways return (countPaths(n - 1, m) + countPaths(n, m - 1)); } // Driver Code let n = 3, m = 2; document.write(" Number of Paths " + countPaths(n, m)); // This code is contributed by shikhasingrajput </script>

**Output**

Number of Paths 10

We can use Dynamic Programming as there are overlapping subproblems. We can draw recursion tree to see overlapping problems. For example, in case of countPaths(4, 4), we compute countPaths(3, 3) multiple times.

## C++

// C++ program to count total number of // paths from a point to origin #include<bits/stdc++.h> using namespace std; // DP based function to count number of paths int countPaths(int n, int m) { int dp[n+1][m+1]; // Fill entries in bottommost row and leftmost // columns for (int i=0; i<=n; i++) dp[i][0] = 1; for (int i=0; i<=m; i++) dp[0][i] = 1; // Fill DP in bottom up manner for (int i=1; i<=n; i++) for (int j=1; j<=m; j++) dp[i][j] = dp[i-1][j] + dp[i][j-1]; return dp[n][m]; } // Driver Code int main() { int n = 3, m = 2; cout << " Number of Paths " << countPaths(n, m); return 0; }

## Java

// Java program to count total number of // paths from a point to origin import java.io.*; class GFG { // DP based function to count number of paths static int countPaths(int n, int m) { int dp[][] = new int[n + 1][m + 1]; // Fill entries in bottommost row and leftmost // columns for (int i = 0; i <= n; i++) dp[i][0] = 1; for (int i = 0; i <= m; i++) dp[0][i] = 1; // Fill DP in bottom up manner for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[n][m]; } // Driver Code public static void main (String[] args) { int n = 3, m = 2; System.out.println(" Number of Paths " + countPaths(n, m)); } } // This code is contributed by vt_m

## Python3

# Python3 program to count total # number of paths from a po to origin # Recursive function to count # number of paths def countPaths(n, m): # If we reach bottom or top # left, we are have only one # way to reach (0, 0) if (n == 0 or m == 0): return 1 # Else count sum of both ways return (countPaths(n - 1, m) + countPaths(n, m - 1)) # Driver Code n = 3 m = 2 print("Number of Paths", countPaths(n, m)) # This code is contributed by ash264

## C#

// C# program to count total number of // paths from a point to origin using System; public class GFG { // DP based function to count number // of paths static int countPaths(int n, int m) { int [,]dp = new int[n + 1,m + 1]; // Fill entries in bottommost row // and leftmost columns for (int i = 0; i <= n; i++) dp[i,0] = 1; for (int i = 0; i <= m; i++) dp[0,i] = 1; // Fill DP in bottom up manner for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) dp[i,j] = dp[i - 1,j] + dp[i,j - 1]; return dp[n,m]; } // Driver Code public static void Main () { int n = 3, m = 2; Console.WriteLine(" Number of" + " Paths " + countPaths(n, m)); } } // This code is contributed by Sam007.

## PHP

<?php // PHP program to count total number of // paths from a point to origin // DP based function to // count number of paths function countPaths($n, $m) { //$dp[$n+1][$m+1]; // Fill entries in bottommost // row and leftmost columns for ($i = 0; $i <= $n; $i++) $dp[$i][0] = 1; for ($i = 0; $i <= $m; $i++) $dp[0][$i] = 1; // Fill DP in bottom up manner for ($i = 1; $i <= $n; $i++) for ($j = 1; $j <= $m; $j++) $dp[$i][$j] = $dp[$i - 1][$j] + $dp[$i][$j - 1]; return $dp[$n][$m]; } // Driver Code $n = 3; $m = 2; echo " Number of Paths " , countPaths($n, $m); // This code is contributed by m_kit ?>

## Javascript

<script> // javascript program to count total number of // paths from a point to origin // DP based function to count number of paths function countPaths(n , m) { var dp = Array(n+1).fill(0).map(x => Array(m+1).fill(0)); // Fill entries in bottommost row and leftmost // columns for (i = 0; i <= n; i++) dp[i][0] = 1; for (i = 0; i <= m; i++) dp[0][i] = 1; // Fill DP in bottom up manner for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[n][m]; } // Driver Code var n = 3, m = 2; document.write(" Number of Paths " + countPaths(n, m)); // This code is contributed by Amit Katiyar </script>

**Output**

Number of Paths 10

**Another Approach:**

Using Pascal’s Triangle Approach, we also solve the problem by calculating the value of ** ^{n+m}C_{n}**. It can be observed as a pattern when you increase the value of

**m**keeping the value of

**n**constant.

Below is the implementation of the above approach:

## C++

// C++ Program for above approach #include <iostream> #include <bits/stdc++.h> using namespace std; // Function to find // binomial Coefficient int binomialCoeff(int n, int k) { int C[k+1]; memset(C, 0, sizeof(C)); C[0] = 1; // Constructing Pascal's Triangle for (int i = 1; i <= n; i++) { for (int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j-1]; } return C[k]; } //Driver Code int main() { int n=3, m=2; cout<<"Number of Paths: "<< binomialCoeff(n+m,n)<<endl; return 0; } //Contributed by Vismay_7

## Java

// Java Program for above approach import java.io.*; import java.util.*; class GFG { static int min(int a,int b) { return a<b?a:b; } // Function for binomial // Coefficient static int binomialCoeff(int n, int k) { int C[] = new int[k + 1]; C[0] = 1; //Constructing Pascal's Triangle for (int i = 1; i <= n; i++) { for (int j = min(i,k); j > 0; j--) C[j] = C[j] + C[j-1]; } return C[k]; } // Driver Code public static void main (String[] args) { int n=3,m=2; System.out.println("Number of Paths: " + binomialCoeff(n+m,n)); } } //Contributed by Vismay_7

## Python3

# Python3 program for above approach def binomialCoeff(n,k): C = [0]*(k+1) C[0] = 1 # Computing Pascal's Triangle for i in range(1, n + 1): j = min(i ,k) while (j > 0): C[j] = C[j] + C[j-1] j -= 1 return C[k] # Driver Code n=3 m=2 print("Number of Paths:",binomialCoeff(n+m,n)) # Contributed by Vismay_7

## C#

// C# program for above approach using System; class GFG{ // Function to find // binomial Coefficient static int binomialCoeff(int n, int k) { int[] C = new int[k + 1]; C[0] = 1; // Constructing Pascal's Triangle for(int i = 1; i <= n; i++) { for(int j = Math.Min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1]; } return C[k]; } // Driver code static void Main() { int n = 3, m = 2; Console.WriteLine("Number of Paths: " + binomialCoeff(n + m, n)); } } // This code is contributed by divyesh072019

## Javascript

<script> // javascript Program for above approach function min(a , b) { return a < b ? a : b; } // Function for binomial // Coefficient function binomialCoeff(n , k) { var C = Array(k + 1).fill(0); C[0] = 1; // Constructing Pascal's Triangle for (i = 1; i <= n; i++) { for (j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1]; } return C[k]; } // Driver Code var n = 3, m = 2; document.write("Number of Paths: " + binomialCoeff(n + m, n)); // This code is contributed by Amit Katiyar </script>

**Output**

Number of Paths: 10

This article is contributed by **Nikhil Rawat**.

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