Q.Let $A = \{1, 2, 3, 4\}$. Write the power set $P(A)$ and verify that $n(P(A)) = 2^4$ by counting the subsets of each possible size separately.

List all subsets grouped by how many elements each contains, then add up the counts from every size group as an independent check against the $2^n$ formula.

Counting subsets of size 0 through 4 separately and adding gives the same total the formula predicts.

Step 1 — List subsets of size 0. Only $\emptyset$ — 1 subset.

Step 2 — List subsets of size 1. $\{1\}, \{2\}, \{3\}, \{4\}$ — 4 subsets.

Step 3 — List subsets of size 2. $\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$ — 6 subsets.

Step 4 — List subsets of size 3. $\{1,2,3\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}$ — 4 subsets.

Step 5 — List subsets of size 4. Only $\{1,2,3,4\}$ itself — 1 subset.

Step 6 — Add up every size group. $1 + 4 + 6 + 4 + 1 = 16$.

Step 7 — Cross-check against the formula. $n(A) = 4$, so $n(P(A)) = 2^4 = 16$ — the size-by-size count and the formula agree exactly, giving

Instead of listing every subset directly, count size-2 subsets alone as a spot-check using the combinations idea: choosing 2 elements out of 4 to include, in any order, gives $^{4}C_{2} = 6$ — matching the direct list of 6 two-element subsets exactly, confirming the size-by-size breakdown independently of listing everything by hand.

Missing one or two subsets while listing by hand, especially in the size-2 group (6 subsets is easy to under-count if not listed systematically, e.g. by always starting from the smallest remaining element). Also, stopping the listing at $A$ has 4 elements → $2^4=16$ subsets without actually verifying it by a genuine independent count — the whole point of this worked example is that the size-by-size total and the formula must be checked against EACH OTHER, not just quoted.