Count of N-digit numbers with all distinct digits

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

Examples:

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

Input: N = 3
Output: 648

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

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 N^th term will be 9 * 9! / (10 – N)!.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
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 

Output:

Time Complexity: O(n)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes

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

C++

Java

Python3

C#

Recommended Posts: