# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

## CPP

 `// 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 `

Output:

```210
```

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Improved By : mohit kumar 29, AnkitRai01