AMC 10 · 2008 · #23
Grade 7 countingTwo subsets of the set S={a,b,c,d,e} are to be chosen so that their union is S and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $S=\{a,b,c,d,e\}$ has $5$ elements; The union of the two chosen subsets is all of $S$; The intersection of the two subsets has exactly $2$ elements; The order in which the subsets are chosen does not matter; Answer choices: (A) $20$, (B) $40$, (C) $60$, (D) $160$, (E) $320$
Unknowns: The number of unordered pairs of subsets meeting both conditions
Understand
Restated: 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.
Givens: $S=\{a,b,c,d,e\}$ has $5$ elements; The union of the two chosen subsets is all of $S$; The intersection of the two subsets has exactly $2$ elements; The order in which the subsets are chosen does not matter; Answer choices: (A) $20$, (B) $40$, (C) $60$, (D) $160$, (E) $320$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #16 Change Focus / Count the Complement
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.
Execute — Answer: B
7.SP.C.8 Step 1 Give each element a role
- Instead of choosing two whole subsets, look at one element at a time.
- Each of the $5$ elements sits in the first subset only, the second subset only, or in both.
- It cannot sit in neither, because the union must cover all of $S$.
- So the two conditions are really rules about how many elements take the "both" role.
💡 Union equals $S$ just means nobody is left out, so every letter must take one of three spots.
7.SP.C.8 Step 2 Choose the two shared elements
- The intersection must have exactly $2$ elements, so exactly $2$ of the $5$ letters take the "both" role.
- Count the ways by listing every pair: $ab, ac, ad, ae, bc, bd, be, cd, ce, de$.
- That is $10$ pairs, which is $\binom{5}{2}=10$.
💡 Picking which $2$ of $5$ letters are shared is just listing all the pairs.
6.EE.A.1 Step 3 Split the other three elements
- The remaining $3$ letters each go into exactly one subset — first-only or second-only.
- They cannot be "both" (that would make the intersection larger than $2$), and they cannot be "neither".
- Each of the $3$ has $2$ choices, and the choices are independent, so there are $2 \times 2 \times 2 = 2^3 = 8$ ways.
💡 Three independent left-or-right forks give $2^3$ endings, like flipping a coin three times.
7.SP.C.8 Step 4 Combine, then fix the double count
- Choosing the shared pair ($10$ ways) and splitting the rest ($8$ ways) are independent, so there are $10 \times 8 = 80$ ways if we treat the subsets as labeled "first" and "second".
- But their order does not matter.
- Every pair is counted twice, once for each labeling, because the two subsets are always different — their overlap is $2$ elements while their union is $5$, so they can never be identical.
- Halving removes the double count: $80 \div 2 = 40$, choice (B).
💡 Labeling the two sets counts each unordered pair twice, so cut the total in half.
7.SP.C.8 Instead of choosing two whole subsets, look at one element at a time. Each of th 7.SP.C.8 The intersection must have exactly $2$ elements, so exactly $2$ of the $5$ lette 6.EE.A.1 The remaining $3$ letters each go into exactly one subset — first-only or second 7.SP.C.8 Choosing the shared pair ($10$ ways) and splitting the rest ($8$ ways) are indep Review
Reasonableness: 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 $\binom{5}{2}=10$ pairs, and sizes $3$ and $4$ give $10 \times 3 = 30$ pairs, totaling $10 + 30 = 40$.
Alternative: Break it into cases by subset size. Since the union has $5$ elements and the overlap has $2$, the two sizes must add to $5 + 2 = 7$, giving sizes $2$ and $5$, or sizes $3$ and $4$. For sizes $2$ and $5$, the small subset is the intersection itself and the big one is all of $S$: $\binom{5}{2}=10$ ways. For sizes $3$ and $4$, pick the $2$ shared elements and then which one of the leftover $3$ joins the smaller subset: $10 \times 3 = 30$ ways. Adding gives $10 + 30 = 40$ — the same choice (B).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the three roles and the $10$ shared pairs, then using the counting principle to combine independent stages and halve away the ordering double count.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Counting the ways to send the three leftover elements to one of two subsets as $2^3 = 8$.)
⭐ 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 — 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.
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