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:
- All the digits of the numbers are from the range [0, N].
- There are no leading 0s.
- 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
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 <bits/stdc++.h> 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.. |
Javascript
<script> // Javascript implementation of the approach // Function to return the factorial of n function fact(n) { let res = 1; for (let i = 2; i <= n; i++) res = res * i; return res; } // Function to return the // count of numbers possible function Count_number(N) { return (N * fact(N)); } // Driver code let N = 2; document.write(Count_number(N)); </script> |
4
Time Complexity: O(n)
Auxiliary Space: O(1)
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