AMC 10 · 2007 · #12
Easy mode Grade 4Two tour guides are splitting up six tourists. Each tourist joins one of the two guides. Neither guide is allowed to end up with no tourists. In how many different ways can the six tourists be split between the two guides?
Pick an answer.
AMC 10 2007 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 tour guides split up six tourists. Each tourist joins one of the two guides, and neither guide is allowed to end up with nobody. Count how many different ways the six tourists can be divided between the two guides.
Givens: There are exactly two guides and six tourists; Every tourist must join exactly one of the two guides; Each guide must end up with at least one tourist; Answer choices: (A) $56$, (B) $58$, (C) $60$, (D) $62$, (E) $64$
Unknowns: The number of allowed ways to divide the six tourists between the two guides
Understand
Restated: Two tour guides split up six tourists. Each tourist joins one of the two guides, and neither guide is allowed to end up with nobody. Count how many different ways the six tourists can be divided between the two guides.
Givens: There are exactly two guides and six tourists; Every tourist must join exactly one of the two guides; Each guide must end up with at least one tourist; Answer choices: (A) $56$, (B) $58$, (C) $60$, (D) $62$, (E) $64$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
The phrase 'each guide must take at least one tourist' is the classic signal for Tool #16 (Count the Complement): instead of directly counting the allowed splits, count every split with no rule at all, then subtract the handful that break the rule. Counting all splits is easy with Tool #7 (Identify Subproblems), because each tourist's choice is its own independent little decision, and independent choices multiply. The only forbidden splits are the two where everybody piles onto a single guide, so removing them is a single subtraction. Tool #2 (Make a Systematic List) gives an independent check by counting the splits by group size.
Execute — Answer: D
3.OA.A.1 Step 1 Count every split, rule ignored
- Forget the 'at least one' rule for a moment.
- Each tourist independently decides between guide A and guide B, so each tourist has $2$ choices.
- Because the six choices are made independently, the number of ways multiplies: $2$ for the first tourist, times $2$ for the second, and so on for all six.
- That is $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{6} = 64$ splits in total.
💡 Every extra tourist who can go two ways doubles the number of possible splits.
4.OA.A.3 Step 2 Find the forbidden splits
- The rule says neither guide can be left empty.
- Among the $64$ splits, which ones break the rule?
- A guide is empty only when every single tourist chose the other guide.
- That happens in exactly two ways: all six tourists go to guide A (leaving B empty), or all six go to guide B (leaving A empty).
- So there are $2$ forbidden splits.
💡 A guide is empty only in the extreme case where the whole group swings to the other guide.
4.OA.A.3 Step 3 Subtract to get the answer
- The allowed splits are all the splits minus the forbidden ones: $64 - 2 = 62$.
- So there are $62$ groupings where both guides get at least one tourist, which is choice (D).
💡 Take away just the two bad cases and everything that remains obeys the rule.
3.OA.A.1 Forget the 'at least one' rule for a moment. Each tourist independently decides 4.OA.A.3 The rule says neither guide can be left empty. Among the $64$ splits, which ones 4.OA.A.3 The allowed splits are all the splits minus the forbidden ones: $64 - 2 = 62$. S Review
Reasonableness: The answer $62$ sits inside the range of the choices ($56$ to $64$) and is just $2$ below the no-rule total of $64$, which fits the fact that only two splits ever break the rule. It is also an even number, as it should be: every valid split can be paired with its mirror image (swap the two guides), so the valid splits come in pairs, forcing an even count.
Alternative: Count the valid splits directly by how many tourists guide A gets. Guide A must get somewhere from $1$ to $5$ tourists (never $0$ or all $6$). Choosing which tourists go to A gives $\binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5} = 6+15+20+15+6 = 62$, matching the complement count.
CCSS standards used (min grade 4)
3.OA.A.1Interpret products of whole numbers as combining equal groups of choices (Reading the six independent two-way choices as a product $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.)4.OA.A.3Solve multistep problems using the four operations (Identifying the two forbidden all-to-one-guide splits and subtracting them, $64 - 2 = 62$.)
⭐ When a rule says each side must get at least one, count every possible split first, then just subtract the few splits that leave a side empty.
⭐ When a rule says each side must get at least one, count every possible split first, then just subtract the few splits that leave a side empty.
More like this
Same archetype — closest grade level first.