1. **Stating the problem:** We want to find the total number of combinations possible from a set of items. 2. **Formula used:** The number of combinations of choosing $k$ items from $n$ items is given by the binomial coefficient: $$ C(n,k) = \frac{n!}{k!(n-k)!} $$ where $n!$ denotes the factorial of $n$. 3. **Important rules:** - Order does not matter in combinations. - $k$ must be between $0$ and $n$ inclusive. 4. **Counting all combinations:** To count all combinations of all possible sizes, we sum over all $k$ from $0$ to $n$: $$ \sum_{k=0}^n C(n,k) = \sum_{k=0}^n \frac{n!}{k!(n-k)!} $$ 5. **Intermediate work:** Using the binomial theorem, we know: $$ \sum_{k=0}^n C(n,k) = 2^n $$ 6. **Explanation:** This means the total number of all possible combinations (including the empty set and the full set) of an $n$-element set is $2^n$. **Final answer:** The total number of all combinations from a set of $n$ elements is: $$ 2^n $$