# All possible numbers of N digits and base B without leading zeros

Given number of digits ‘N’ and base ‘B’, the task is to count all the ‘N’ digit numbers without leading zeros that are in base ‘B’.

**Examples:**

Input:N = 2, B = 2Output:2 All possible numbers without leading zeros are 10 and 11.Input:N = 5, B = 8Output:28672

**Approach:**

- If the base is ‘B’ then every digit of the number can take any value within the range [0, B-1].
- So, B ‘N’ digit numbers are possible with base ‘B’ (including the numbers with leading zeros).
- And, if we fix the first digit as ‘0’ then the rest of the ‘N-1’ digits can form a total of B numbers.
- So, total number of ‘N’ digit numbers with base ‘B’ possible without leading zeros are B – B.

Below is the implementation of above Approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to count ` `// all permutations ` `void` `countPermutations(` `int` `N, ` `int` `B) ` `{ ` ` ` `// count of ` ` ` `// all permutations ` ` ` `int` `x = ` `pow` `(B, N); ` ` ` ` ` `// count of permutations ` ` ` `// with leading zeros ` ` ` `int` `y = ` `pow` `(B, N - 1); ` ` ` ` ` `// Return the permutations ` ` ` `// without leading zeros ` ` ` `cout << x - y << ` `"\n"` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ` `int` `N = 6; ` ` ` `int` `B = 4; ` ` ` ` ` `countPermutations(N, B); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` ` ` `class` `GFG ` `{ ` `// function to count ` `// all permutations ` `static` `void` `countPermutations(` `int` `N, ` `int` `B) ` `{ ` ` ` `// count of ` ` ` `// all permutations ` ` ` `int` `x = (` `int` `)Math.pow(B, N); ` ` ` ` ` `// count of permutations ` ` ` `// with leading zeros ` ` ` `int` `y = (` `int` `)Math.pow(B, N - ` `1` `); ` ` ` ` ` `// Return the permutations ` ` ` `// without leading zeros ` ` ` `System.out.println(x - y); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `N = ` `6` `; ` ` ` `int` `B = ` `4` `; ` ` ` ` ` `countPermutations(N, B); ` `} ` `} ` ` ` `// This code is contributed by mits ` |

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## Python3

# Python3 implementation of the approach

# function to count all permutations

def countPermutations(N, B):

# count of all permutations

x = B ** N

# count of permutations

# with leading zeros

y = B ** (N – 1)

# Return the permutations

# without leading zeros

print(x – y)

# Driver code

if __name__ == “__main__”:

N, B = 6, 4

countPermutations(N, B)

# This code is contributed by Rituraj Jain

## C#

`// C# implementation of the approach ` ` ` `using` `System; ` `class` `GFG ` `{ ` `// function to count ` `// all permutations ` `static` `void` `countPermutations(` `int` `N, ` `int` `B) ` `{ ` ` ` `// count of ` ` ` `// all permutations ` ` ` `int` `x = (` `int` `)Math.Pow(B, N); ` ` ` ` ` `// count of permutations ` ` ` `// with leading zeros ` ` ` `int` `y = (` `int` `)Math.Pow(B, N - 1); ` ` ` ` ` `// Return the permutations ` ` ` `// without leading zeros ` ` ` `Console.WriteLine(x - y); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `N = 6; ` ` ` `int` `B = 4; ` ` ` ` ` `countPermutations(N, B); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai(Abby_akku) ` |

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## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// function to count all permutations ` `function` `countPermutations(` `$N` `, ` `$B` `) ` `{ ` ` ` `// count of all permutations ` ` ` `$x` `= pow(` `$B` `, ` `$N` `); ` ` ` ` ` `// count of permutations ` ` ` `// with leading zeros ` ` ` `$y` `= pow(` `$B` `, ` `$N` `- 1); ` ` ` ` ` `// Return the permutations ` ` ` `// without leading zeros ` ` ` `echo` `(` `$x` `- ` `$y` `), ` `"\n"` `; ` `} ` ` ` `// Driver code ` `$N` `= 6; ` `$B` `= 4; ` ` ` `countPermutations(` `$N` `, ` `$B` `); ` ` ` `// This code is contributed ` `// by Sach_Code` ` `?> ` |

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**Output:**

3072

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