AMC 10 · 2008 · #23

Grade 7 counting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

combinations-basicset-partitionsystematic-enumeration caseworkcomplementary-counting ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 3 insights

Problem

Two subsets of S={a,b,c,d,e} are chosen so that their union is all of S and their intersection has exactly two elements. The order of the two subsets does not matter. Find how many such unordered pairs of subsets are possible.

Pick an answer.

(A)

$\ 20$

(B)

$\ 40$

(C)

$\ 60$

(D)

$\ 160$

(E)

$\ 320$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Picking two whole subsets at once is hard to count directly, so Tool #7 (Identify Subproblems) breaks the job into clean stages: choose which elements are shared, then decide the rest, then correct for order. Tool #16 (Change Focus) does the key reframe — instead of choosing sets, look at one element at a time and ask which of three roles it takes (first-only, second-only, or both), because the union rule forbids the fourth role of belonging to neither. Tool #2 (Make a Systematic List) handles each counting stage by listing outcomes, so the multiplication of independent choices is trustworthy rather than guessed.

1STEP 1

Give each element a role

Look at one letter at a time: each of the 5 is first-only, second-only, or both — never neither, since the union covers all of S.

role of each element ∈ {first only, second only, both}

2STEP 2

Choose the two shared elements

Exactly 2 of the 5 letters take the 'both' role; listing every pair ab, ac, ad, ae, bc, bd, be, cd, ce, de gives 10 ways.

C(5, 2) = 10

3STEP 3

Split the other three elements

Each of the other 3 letters must pick one side, first-only or second-only, so 2 × 2 × 2 = 8 ways.

2³ = 8

4STEP 4

Combine, then fix the double count

Labeled, that is 10 × 8 = 80; the two subsets always differ, so each pair got counted twice — halve to get 40, choice (B).

(10 × 8)/2 = 80/2 = 40 (B)

Answer

The answer 40 sits exactly halfway between the small trap 20 and the labeled total 80; halving 80 lands on 40, choice (B). Forgetting to restrict the three leftover letters — letting all elements split freely — overshoots toward the big traps 160 and 320, so those are the careless answers. A size-case check agrees: subsets of sizes 2 and 5 give C(5, 2)=10 pairs, and sizes 3 and 4 give 10 × 3 = 30 pairs, totaling 10 + 30 = 40.

💡Key takeaway

Give each element a role — only-first, only-second, or both — pin exactly two to "both", let the other three pick a side, then halve because the two subsets have no order.

Give each element a role
Choose the two shared elements
Split the other three elements
Combine, then fix the double count

Give each element a role

Choose the two shared elements

Split the other three elements

Combine, then fix the double count

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