Original question: Let a sample space $S$ have $n$ elements. Prove that the number of subsets of $ S $ is $2^n$.
Key insight: The amount of subsets of a set can be counted using combinations. For example, if a set has $ 2 $ elements, then the number of subsets is:
$$ \binom{2}{0} + \binom{2}{1} + \binom{2}{2} = 1 + 2 + 1 = 4 $$
Therefore, if a set $ S $ has $ n $ elements, then $ S $ has
$$ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + ... + \binom{n}{n} $$ subsets
Thus, proving that a set with $ n $ elements has $ 2^n $ subsets amounts to proving the following statement:
$$ \sum_{i = 0}^{n} \binom{n}{i} = 2^n $$
The most obvious way to prove this equality seems to be by induction. A set with $ n = 1 $ elements has $ \binom{1}{0} + \binom{1}{1} = 2 $ subsets, so the base case holds. Now we assume the equality works for $ n $ and show that it works for $ n+1 $:
$$ \sum_{i = 0}^{n+1} \binom{n+1}{i} = 2^{n+1} $$
$$ \sum_{i = 0}^{n} \binom{n+2}{i} = 2 \times 2^n $$ This is where I'm stumbling.
