Number of pairs of arrays (A, B) such that A is ascending, B is descending and A[i] ≤ B[i]

Given two integers N and M, the task is to find the number of pairs of arrays (A, B) such that array A and B both are of size M each where each entry of A and B is an integer between 1 and N such that for each i between 1 and M, A[i] ≤ B[i]. It is also given that the array A is sorted in non-descending order and B is sorted in non-ascending order. Since the answer can be very large, return answer modulo 109 + 7.

Examples:

Input: N = 2, M = 2
Output: 5
1: A= [1, 1] B=[1, 1]
2: A= [1, 1] B=[1, 2]
3: A= [1, 1] B=[2, 2]
4: A= [1, 2] B=[2, 2]
5: A= [2, 2] B=[2, 2]

Input: N = 5, M = 3
Output: 210

Approach: Notice that if there is a valid pair of arrays A and B and if B is concatenated after A the resultant array will always be either an ascending or a non-descending array of size of 2 * M. Each element of (A + B) will be between 1 and N (It is not necessary that all elements between 1 and N have to be used). This now simply converts the given problem to finding all the possible combinations of size 2 * M where each element is between 1 to N (with repetitions allowed) whose formula is 2 * M + N – 1CN – 1 or (2 * M + N – 1)! / ((2 * M)! * (N – 1)!).



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ code of above approach
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
  
long long fact(long long n)
{
    if(n == 1) 
        return 1;
    else
        return (fact(n - 1) * n) % mod;
}
  
// Function to return the count of pairs
long long countPairs(int m, int n)
{
    long long ans = fact(2 * m + n - 1) / 
                    (fact(n - 1) * fact(2 * m));
    return (ans % mod);
}
  
// Driver code
int main()
{
    int n = 5, m = 3;
    cout << (countPairs(m, n));
    return 0;
}
  
// This code is contributed by mohit kumar 29
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java code of above approach 
class GFG 
{
    final static long mod = 1000000007 ;
  
    static long fact(long n) 
    
        if(n == 1
            return 1
        else
            return (fact(n - 1) * n) % mod; 
    
      
    // Function to return the count of pairs 
    static long countPairs(int m, int n) 
    
        long ans = fact(2 * m + n - 1) / 
                   (fact(n - 1) * fact(2 * m)); 
          
        return (ans % mod); 
    
      
    // Driver code 
    public static void main (String[] args)
    
        int n = 5, m = 3
          
        System.out.println(countPairs(m, n)); 
    
}
  
// This code is contributed by AnkitRai01
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
from math import factorial as fact
  
# Function to return the count of pairs
def countPairs(m, n):
    ans = fact(2 * m + n-1)//(fact(n-1)*fact(2 * m))
    return (ans %(10**9 + 7))
  
# Driver code
n, m = 5, 3
print(countPairs(m, n))
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# code of above approach 
using System;
  
class GFG 
{
    static long mod = 1000000007 ;
  
    static long fact(long n) 
    
        if(n == 1) 
            return 1; 
        else
            return (fact(n - 1) * n) % mod; 
    
      
    // Function to return the count of pairs 
    static long countPairs(int m, int n) 
    
        long ans = fact(2 * m + n - 1) / 
                (fact(n - 1) * fact(2 * m)); 
          
        return (ans % mod); 
    
      
    // Driver code 
    public static void Main()
    
        int n = 5, m = 3; 
          
        Console.WriteLine(countPairs(m, n)); 
    
}
  
// This code is contributed by AnkitRai01
chevron_right

Output:
210

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29, AnkitRai01

Article Tags :