# Count of N digit numbers possible which satisfy the given conditions

Given an integer N, the Task is to find the total number of N digit numbers possible such that:

1. All the digits of the numbers are from the range [0, N].
2. There are no leading 0s.
3. All the digits in a number are distinct.

Examples:

Input: N = 2
Output: 4
10, 12, 20 and 21 are the only possible 2 digit
numbers which satisfy the given conditions.

Input: N = 5
Output: 600

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

Approach: Given N number of digit and the first place can be filled in N ways [0 cannot be taken as the first digit and the allowed digits are from the range [1, N]]
Remaining (N – 1) places can be filled in N! ways
So, total count of number possible will be N * N!.

Take an example for better understanding. Say, N = 8 First place can be filled with any digit from [1, 8] and the remaining 7 places can be filled in 8! ways i.e 8 * 7 * 6 * 5 * 4 * 3 * 2.
So, total ways = 8 * 8! = 8 * 8 * 7 * 6 * 5 * 4 * 3 * 2 = 322560

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the factorial of n ` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function to return the ` `// count of numbers possible ` `int` `Count_number(``int` `N) ` `{ ` `    ``return` `(N * fact(N)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 2; ` ` `  `    ``cout << Count_number(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the factorial of n ` `static` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = ``1``; ` `    ``for` `(``int` `i = ``2``; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function to return the ` `// count of numbers possible ` `static` `int` `Count_number(``int` `N) ` `{ ` `    ``return` `(N * fact(N)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `N = ``2``; ` ` `  `    ``System.out.print(Count_number(N)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the factorial of n ` `def` `fact(n): ` ` `  `    ``res ``=` `1` `    ``for` `i ``in` `range``(``2``, n ``+` `1``): ` `        ``res ``=` `res ``*` `i ` `    ``return` `res ` ` `  `# Function to return the ` `# count of numbers possible ` `def` `Count_number(N): ` `    ``return` `(N ``*` `fact(N)) ` ` `  `# Driver code ` `N ``=` `2` ` `  `print``(Count_number(N)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the factorial of n ` `static` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = 1; ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function to return the ` `// count of numbers possible ` `static` `int` `Count_number(``int` `N) ` `{ ` `    ``return` `(N * fact(N)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `N = 2; ` ` `  `    ``Console.WriteLine(Count_number(N)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```4
```

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Improved By : mohit kumar 29, vt_m

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