# Number of ways to reach (M, N) in a matrix starting from the origin without visiting (X, Y)

Last Updated : 06 Sep, 2022

Given four positive integers M, N, X, and Y, the task is to count all the possible ways to reach from the top left(i.e., (0, 0)) to the bottom right (M, N) of a matrix of size (M+1)x(N+1) without visiting the cell (X, Y). It is given that from each cell (i, j) you can either move only to right (i, j + 1) or down (i + 1, j).

Examples:

Input: M = 2, N = 2, X = 1, Y = 1
Output:
Explanation:
There are only 2 ways to reach (2, 2) without visiting (1, 1) and the two paths are:
(0, 0) -> (0, 1) -> (0, 2) -> (1, 2) -> (2, 2)
(0, 0) -> (1, 0) -> (2, 0) -> (2, 1) -> (2, 2)

Input: M = 5, N = 4, X = 3, Y = 2
Output: 66
Explanation:
There are 66 ways to reach (5, 4) without visiting (3, 2).

Approach:
To solve the problem mentioned above the idea is to subtract the number of ways to reach from (0, 0) to (X, Y) which was followed by reaching (M, N) from (X, Y) by visiting (X, Y) from the total number of ways reaching (M, N) from (0, 0)
Therefore,

1. The number of ways to reach from (M, N) from the origin (0, 0) is given by:

1. The number of ways to reach (M, N) only by visiting (X, Y) is reaching (X, Y) from (0, 0) which was followed by reaching (M, N) from (X, Y) is given by:

Therefore,

1. Hence, the equation for the total number of ways are:

Below is the implementation of the above approach:

## C++

 // C++ program from the above approach#include using namespace std; int fact(int n); // Function for computing nCrint nCr(int n, int r){    return fact(n)           / (fact(r) * fact(n - r));} // Function to find factorial of a numberint fact(int n){    int res = 1;     for (int i = 2; i <= n; i++)        res = res * i;     return res;} // Function for counting the number// of ways to reach (m, n) without// visiting (x, y)int countWays(int m, int n, int x, int y){    return nCr(m + n, m)           - nCr(x + y, x) * nCr(m + n                                     - x - y,                                 m - x);} // Driver Codeint main(){    // Given Dimensions of Matrix    int m = 5;    int n = 4;     // Cell not to be visited    int x = 3;    int y = 2;     // Function Call    cout << countWays(m, n, x, y);    return 0;}

## Java

 // Java program from the above approach     import java.util.*;     class GFG{         // Function for computing nCr     public static int nCr(int n, int r)         {         return fact(n) / (fact(r) * fact(n - r));         }              // Function to find factorial of a number     public static int fact(int n)     {         int res = 1;         for(int i = 2; i <= n; i++)                 res = res * i;             return res;         }              // Function for counting the number         // of ways to reach (m, n) without         // visiting (x, y)         public static int countWays(int m, int n,                            int x, int y)         {         return nCr(m + n, m) -            nCr(x + y, x) *            nCr(m + n - x - y, m - x);         }  // Driver codepublic static void main(String[] args){              // Given Dimensions of Matrix         int m = 5;             int n = 4;                          // Cell not to be visited         int x = 3;             int y = 2;                          // Function Call         System.out.println(countWays(m, n, x, y));     }     } // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 program for the above approach # Function for computing nCrdef nCr(n, r):         return (fact(n) // (fact(r) *                        fact(n - r))) # Function to find factorial of a numberdef fact(n):         res = 1    for i in range(2, n + 1):        res = res * i     return res # Function for counting the number# of ways to reach (m, n) without# visiting (x, y)def countWays(m, n, x, y):         return (nCr(m + n, m) - nCr(x + y, x) *            nCr(m + n - x - y, m - x)) # Driver Code # Given dimensions of Matrixm = 5n = 4 # Cell not to be visitedx = 3y = 2 # Function callprint(countWays(m, n, x, y)) # This code is contributed by sanjoy_62

## C#

 // C# program from the above approach     using System; class GFG{      // Function for computing nCr     public static int nCr(int n, int r)         {         return fact(n) / (fact(r) * fact(n - r));         }              // Function to find factorial of a number     public static int fact(int n)     {         int res = 1;         for(int i = 2; i <= n; i++)                 res = res * i;             return res;         }              // Function for counting the number         // of ways to reach (m, n) without         // visiting (x, y)         public static int countWays(int m, int n,                            int x, int y)         {         return nCr(m + n, m) -            nCr(x + y, x) *            nCr(m + n - x - y, m - x);         }  // Driver codepublic static void Main(String[] args){              // Given dimensions of Matrix         int m = 5;             int n = 4;                          // Cell not to be visited         int x = 3;             int y = 2;                          // Function call         Console.WriteLine(countWays(m, n, x, y));     }     } // This code is contributed by Rajput-Ji

## Javascript

 

Output
66

Time Complexity: O(M + N), where M, N represents the size of the matrix.
Auxiliary Space: O(1), as constant space is required.