Related Articles

# Count of N-digit numbers with all distinct digits

• Difficulty Level : Hard
• Last Updated : 01 Apr, 2021

Given an integer N, the task is to find the count of N-digit numbers with all distinct digits.
Examples:

Input: N = 1
Output:
1, 2, 3, 4, 5, 6, 7, 8 and 9 are the 1-digit numbers
with all distinct digits.
Input: N = 3
Output: 648

Approach: If N > 10 i.e. there will be atleast one digit which will be repeating hence for such cases the answer will be 0 else for the values of N = 1, 2, 3, …, 9, a series will be formed as 9, 81, 648, 4536, 27216, 136080, 544320, … whose Nth term will be 9 * 9! / (10 – N)!.
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` `factorial(``int` `n)``{``    ``if` `(n == 0)``        ``return` `1;``    ``return` `n * factorial(n - 1);``}` `// Function to return the count``// of n-digit numbers with``// all distinct digits``int` `countNum(``int` `n)``{``    ``if` `(n > 10)``        ``return` `0;``    ``return` `(9 * factorial(9)``            ``/ factorial(10 - n));``}` `// Driver code``int` `main()``{``    ``int` `n = 3;` `    ``cout << countNum(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to return the factorial of n``    ``static` `int` `factorial(``int` `n)``    ``{``        ``if` `(n == ``0``)``            ``return` `1``;``        ``return` `n * factorial(n - ``1``);``    ``}``    ` `    ``// Function to return the count``    ``// of n-digit numbers with``    ``// all distinct digits``    ``static` `int` `countNum(``int` `n)``    ``{``        ``if` `(n > ``10``)``            ``return` `0``;``        ``return` `(``9` `* factorial(``9``) /``                    ``factorial(``10` `- n));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String []args)``    ``{``        ``int` `n = ``3``;``        ``System.out.println(countNum(n));``    ``}``}` `// This code is contributed by Srathore`

## Python3

 `# Python3 implementation of the approach` `# Function to return the factorial of n``def` `factorial(n) :` `    ``if` `(n ``=``=` `0``) :``        ``return` `1``;``    ``return` `n ``*` `factorial(n ``-` `1``);` `# Function to return the count``# of n-digit numbers with``# all distinct digits``def` `countNum(n) :``    ``if` `(n > ``10``) :``        ``return` `0``;``        ` `    ``return` `(``9` `*` `factorial(``9``) ``/``/``                ``factorial(``10` `-` `n));` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `3``;` `    ``print``(countNum(n));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``                    ` `class` `GFG``{``    ` `    ``// Function to return the factorial of n``    ``static` `int` `factorial(``int` `n)``    ``{``        ``if` `(n == 0)``            ``return` `1;``        ``return` `n * factorial(n - 1);``    ``}``    ` `    ``// Function to return the count``    ``// of n-digit numbers with``    ``// all distinct digits``    ``static` `int` `countNum(``int` `n)``    ``{``        ``if` `(n > 10)``            ``return` `0;``        ``return` `(9 * factorial(9) /``                    ``factorial(10 - n));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``int` `n = 3;``        ``Console.WriteLine(countNum(n));``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:

`648`

Time Complexity: O(n)

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up