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

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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: 2
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,

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

$\text{Total ways from (0, 0) to (M, N)} = \binom{M + N}{M}$

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:

$\text{Total ways from (0, 0) to (X, Y)} = \binom{X + Y}{X}$
$\text{Total ways from (X, Y) to (M, N)} = \binom{M + N - X - Y}{M - X}$

Therefore,

$\text{Total ways from (0, 0) to (M, N) only by visting (X, Y)} = (\binom{X + Y}{X}) * (\binom{M + N - X - Y}{M - X})$

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

$\binom{M + N}{M} - (\binom{X + Y}{X}) * (\binom{M + N - X - Y}{M - X})$

Below is the implementation of the above approach:

C++

// C++ program from the above approach
#include <bits/stdc++.h>
using namespace std;
 
int fact(int n);
 
// Function for computing nCr
int nCr(int n, int r)
{
    return fact(n)
           / (fact(r) * fact(n - r));
}
 
// Function to find factorial of a number
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)
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 Code
int 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 code
public 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 nCr
def nCr(n, r):
     
    return (fact(n) // (fact(r) *
                        fact(n - r)))
 
# Function to find factorial of a number
def 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 Matrix
m = 5
n = 4
 
# Cell not to be visited
x = 3
y = 2
 
# Function call
print(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 code
public 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

<script>
 
 
// Javascript Program to implement
// the above approach
 
// Function for computing nCr     
function nCr(n, r)         
{     
    return fact(n) / (fact(r) * fact(n - r));         
}     
           
// Function to find factorial of a number     
function fact(n)     
{     
    let res = 1;
       
    for(let 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)         
function 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 Matrix     
    let m = 5;         
    let n = 4;         
               
    // Cell not to be visited     
    let x = 3;         
    let y = 2;         
               
    // Function Call     
    document.write(countWays(m, n, x, y));
 
// This code is contributed by avijitmondal1998.
</script>

Output

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

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