Three-State Subset Count with Bold Elements
Erica is constructing a subset $S$ of $[n] = \{1, 2, \ldots, n\}$. For any element $x \in [n]$, she has three options:
- Exclude $x$ from $S$
- Include $x$ in $S$ (confident)
- Include $x$ in $S$ but boldface it (uncertain)
An element may appear at most once in $S$, so the set $\{1, 2, \mathbf{2}, 4\}$ is invalid. For example, $\{1, 2\}$, $\{1, \mathbf{3}, 4\}$, and $\{\mathbf{1}, \mathbf{2}, \mathbf{3}, \mathbf{4}\}$ are all valid.
How many valid subsets $S$ can Erica construct when $n = 5$?
Hints
- Think about each element's decision independently -- how many choices does Erica have for a single element $x$?
- Each element has exactly 3 options: exclude, include normally, or include in bold. These choices are mutually exclusive and independent across elements.
- Apply the multiplication principle: with 3 independent choices for each of 5 elements, the total count is $3^5$.
Worked Solution
How to Think About It: The trick is recognizing that each element is handled independently. For any single element, there are exactly three mutually exclusive states: out, in (confident), in (bold). Once you see that, the counting is immediate -- it is just $3^n$.
Quick Estimate: With $n = 5$ and 3 choices per element: $3^5 = 243$. Sanity check: ordinary subsets (without bold) give $2^5 = 32$. Adding a third option per element multiplies the count by another factor: $32 \times (3/2)^5 \approx 32 \times 7.59 \approx 243$. Checks out.
Formal Solution:
For each element $x \in \{1, 2, 3, 4, 5\}$, there are exactly 3 choices: 1. Exclude $x$ (not in $S$) 2. Include $x$ normally 3. Include $x$ in bold
These choices are independent across elements, so by the multiplication principle: $$|\mathcal{S}| = 3^n$$
For $n = 5$: $$3^5 = 243$$
Answer: $\boxed{243}$
Intuition
This is a combinatorics counting problem dressed up in unusual notation. The bold/non-bold distinction adds a third state per element, turning a binary (in/out) problem into a ternary one. Recognizing this immediately collapses the problem to $3^n$.
The broader pattern -- counting objects where each component has $k$ independent states -- appears constantly in quant interviews. Binary strings of length $n$ have $2^n$ possibilities; ternary strings have $3^n$; the number of functions from a set of size $n$ to a set of size $k$ is $k^n$. Any time you can decompose a combinatorial object into independent per-element decisions, the total count is the product of per-element counts.