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 
 

Naive 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++ 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 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 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# 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

<script>


// Javascript implementation of the approach

// Function to return the factorial of n
function 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
function countNum(n)
{
    if (n > 10)
        return 0;
    return (9 * factorial(9)
            / factorial(10 - n));
}

// Driver code
var n = 3;
document.write(countNum(n));

// This code is contributed by rutvik_56.
</script>

648

Time Complexity: O(n)

Auxiliary Space: O(n)

Efficient Approach:  We have to fill (n) places with different digit. Like we n=2 then (_ _) places to fill. first place we fill (1 to 9) any number. let we fill 9 in first place then in second place we have choice (0 to 8). So for first place we 9 choices because we can not fill 0 at first place and after that for 2nd place we 9 choice and for 3rd place we 8 choice then so on... 

let we take 4 digit no so ( 9choice      9 choice      8  choice       7choice) .

Choices->

first place -9

second place-9

third place -8

fourth place -7

and so on.......

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of n-digit numbers with
// all distinct digits
long long countNum(int n)
{
    if (n > 10)
        return 0;

    long long count = 1; // Store the count
    long long j = 9; // take choice

    /* take loop 1 to n and multiply with choice*/
    for (int i = 1; i <= n; i++) {

        if (i == 1) {

            count = count * j;
            continue;
        }
        else {
            count = count * j;
            j--;
        }
    }
  return count;
}

// Driver code
int main()
{
    int n = 3;

    cout << countNum(n);

    return 0;
}

// Java implementation of the approach
class GFG {

    // Function to return the count
    // of n-digit numbers with
    // all distinct digits
    static long countNum(int n)
    {
        if (n > 10)
            return 0;
        long count = 1; // Store the count
        long j = 9; // take choice

        /* take loop 1 to n and multiply with choice*/
        for (int i = 1; i <= n; i++) {

            if (i == 1) {

                count = count * j;
                continue;
            }
            else {
                count = count * j;
                j--;
            }
        }
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 3;
        System.out.println(countNum(n));
    }
}

// This code is contributed by Srathore

# Python implementation of the approach

# Function to return the count
# of n-digit numbers with
# all distinct digits
def countNum(n):
    if n > 10:
        return 0
    count = 1  # Store the count
    j = 9  # take choice

    # take loop 1 to n and multiply with choice
    for i in range(1, n+1):
        if i == 1:
            count = count * j
            continue
        else:
            count = count * j
            j -= 1
    return count


# Driver code
n = 3
print(countNum(n))

# This code is contributed by rutikbhosale

using System;

class GFG
{

  // Function to return the count
  // of n-digit numbers with
  // all distinct digits
  static long countNum(int n)
  {
    if (n > 10)
      return 0;
    long count = 1; // Store the count
    long j = 9; // take choice

    /* take loop 1 to n and multiply with choice*/
    for (int i = 1; i <= n; i++)
    {

      if (i == 1)
      {

        count = count * j;
        continue;
      }
      else
      {
        count = count * j;
        j--;
      }
    }
    return count;
  }

  // Driver code
  static void Main(string[] args)
  {
    int n = 3;
    Console.WriteLine(countNum(n));
  }
}

// Javascript implementation of the approach

// Function to return the count
// of n-digit numbers with
// all distinct digits
function countNum(n) {
  if (n > 10)
    return 0;

  let count = 1; // Store the count
  let j = 9; // take choice

  /* take loop 1 to n and multiply with choice*/
  for (let i = 1; i <= n; i++) {
    if (i === 1) {
      count *= j;
      continue;
    } else {
      count *= j;
      j--;
    }
  }
  return count;
}

// Driver code
const n = 3;
console.log(countNum(n));

648

Time Complexity: O(n)

Auxiliary Space: O(1)