# Count of permutations of an Array having each element as a multiple or a factor of its index

Given an integer, N, the task is to count the number of ways to generate an array, arr[] of consisting of N integers such that for every index i(1-based indexing), arr[i] is either a factor or a multiple of i, or both. The arr[] must be the permutations of all the numbers from the range [1, N].

Examples:

Input: N=2
Output: 2
Explanation:
Two possible arrangements are {1, 2} and {2, 1}
Input: N=3
Output: 3
Explanation:
The 6 possible arrangements are {1, 2, 3}, {2, 1, 3}, {3, 2, 1}, {3, 1, 2}, {2, 3, 1} and {1, 3, 2}.
Among them, the valid arrangements are {1, 2, 3}, {2, 1, 3} and {3, 2, 1}.

Approach: The problem can be solved using Backtracking technique and the concept of print all permutations using recursion. Follow the steps below to find the recurrence relation:

1. Traverse the range [1, N].
2. For the current index pos, if i % pos == 0 and i % pos == 0, then insert i into the arrangement and use the concept of Backtracking to find valid permutations.
3. Remove i.
4. Repeat the above steps for all values in the range [1, N] and finally, print the count of valid permutations.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the count of` `// desired permutations` `int` `findPermutation(unordered_set<``int``>& arr,` `                    ``int` `N)` `{` `    ``int` `pos = arr.size() + 1;`   `    ``// Base case` `    ``if` `(pos > N)` `        ``return` `1;`   `    ``int` `res = 0;`   `    ``for` `(``int` `i = 1; i <= N; i++) {`   `        ``// If i has not been inserted` `        ``if` `(arr.find(i) == arr.end()) {`   `            ``// Backtrack` `            ``if` `(i % pos == 0 or pos % i == 0) {`   `                ``// Insert i` `                ``arr.insert(i);`   `                ``// Recur to find valid permutations` `                ``res += findPermutation(arr, N);`   `                ``// Remove i` `                ``arr.erase(arr.find(i));` `            ``}` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `res;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 5;` `    ``unordered_set<``int``> arr;` `    ``cout << findPermutation(arr, N);`   `    ``return` `0;` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.util.*;`   `class` `GFG{` `    `  `// Function to find the count of` `// desired permutations` `static` `int` `findPermutation(Setarr,` `                           ``int` `N)` `{` `    ``int` `pos = arr.size() + ``1``;`   `    ``// Base case` `    ``if` `(pos > N)` `        ``return` `1``;`   `    ``int` `res = ``0``;`   `    ``for``(``int` `i = ``1``; i <= N; i++) ` `    ``{` `        `  `        ``// If i has not been inserted` `        ``if` `(! arr.contains(i))` `        ``{` `            `  `            ``// Backtrack` `            ``if` `(i % pos == ``0` `|| pos % i == ``0``)` `            ``{` `                `  `                ``// Insert i` `                ``arr.add(i);`   `                ``// Recur to find valid permutations` `                ``res += findPermutation(arr, N);`   `                ``// Remove i` `                ``arr.remove(i);` `            ``}` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `res;` `}`   `// Driver Code` `public` `static` `void` `main(String []args)` `{` `    ``int` `N = ``5``;` `    ``Set arr = ``new` `HashSet();` `    `  `    ``System.out.print(findPermutation(arr, N));` `}` `}`   `// This code is contributed by chitranayal`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to find the count of` `# desired permutations` `def` `findPermutation(arr, N):`   `    ``pos ``=` `len``(arr) ``+` `1`   `    ``# Base case` `    ``if``(pos > N):` `        ``return` `1`   `    ``res ``=` `0`   `    ``for` `i ``in` `range``(``1``, N ``+` `1``):`   `        ``# If i has not been inserted` `        ``if``(i ``not` `in` `arr):`   `            ``# Backtrack` `            ``if``(i ``%` `pos ``=``=` `0` `or` `pos ``%` `i ``=``=` `0``):`   `                ``# Insert i` `                ``arr.add(i)`   `                ``# Recur to find valid permutations` `                ``res ``+``=` `findPermutation(arr, N)`   `                ``# Remove i` `                ``arr.remove(i)`   `    ``# Return the final count` `    ``return` `res`   `# Driver Code` `N ``=` `5` `arr ``=` `set``()`   `# Function call` `print``(findPermutation(arr, N))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{` `    `  `// Function to find the count of` `// desired permutations` `static` `int` `findPermutation(HashSet<``int``>arr,` `                           ``int` `N)` `{` `    ``int` `pos = arr.Count + 1;`   `    ``// Base case` `    ``if` `(pos > N)` `        ``return` `1;`   `    ``int` `res = 0;`   `    ``for``(``int` `i = 1; i <= N; i++) ` `    ``{` `        `  `        ``// If i has not been inserted` `        ``if` `(! arr.Contains(i))` `        ``{` `            `  `            ``// Backtrack` `            ``if` `(i % pos == 0 || pos % i == 0)` `            ``{` `                `  `                ``// Insert i` `                ``arr.Add(i);`   `                ``// Recur to find valid permutations` `                ``res += findPermutation(arr, N);`   `                ``// Remove i` `                ``arr.Remove(i);` `            ``}` `        ``}` `    ``}`   `    ``// Return the readonly count` `    ``return` `res;` `}`   `// Driver Code` `public` `static` `void` `Main(String []args)` `{` `    ``int` `N = 5;` `    ``HashSet<``int``> arr = ``new` `HashSet<``int``>();` `    `  `    ``Console.Write(findPermutation(arr, N));` `}` `}`   `// This code is contributed by gauravrajput1`

Output:

```10

```

Time Complexity: O(N×N!)
Auxiliary Space: O(N)

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Improved By : chitranayal, GauravRajput1