🎯 Derivation through Combinatorics

Every place value has choices ranging from 0 to the base number b-1, which is b choices

$$ 0, \space 1, \space 2, \space 3, \cdots, \space b-2, \space b-1 $$

And with n places, we have a combined outcome of n outcomes

$$ \underline{\phantom{()} \text{n}^{th} \space \text{place}} \quad \underline{\text{n-1}^{th} \space \text{place}} \quad \cdots \quad \underline{\text{2}^{nd} \space \text{place}} \quad \underline{\text{1}^{st} \space \text{place}} $$

Since each outcome there are b choices, and we have n outcomes, totalling to $b^n$ outcomes

$$ b^n = \underbrace{b \times b \times b \times \cdots \times b \times b}_{\text{n times}} $$

We have a total of $b^n$ possible numbers

🎯 Derivation through Number System

Counting from 1 to the maximum value of the base b for n digits

$$ (b-1)b^{n-1} + (b-1)b^{n-2} + \cdots + (b-1)b^1 + (b-1)b^0 $$

Which is also the same as the place value of the next place minus 1

$$ b^n - 1 $$

Then we count 0 as the unaccounted representation

$$ (b^n - 1) + 1 $$

We get

$$ b^n $$