# 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:
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:

## C++

 `// C++ code of above approach` `#include ` `#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`

## Java

 `// 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`

## Python3

 `# 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))`

## C#

 `// 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`

## Javascript

 ``

Output:

`210`

Time Complexity: O(n + m)
Auxiliary Space: O(max(n, m)).

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next