AMC 10 · 2007 · #12

Grade 4 arithmetic
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
complementary-countingcombinations-basicexponents complementary-counting ↑ Prerequisites: complementary-counting
📏 Medium solution 💡 2 insights
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Problem
Two tour guides are leading six tourists. The guides decide to split up, and each tourist must choose one of the two guides. There is one rule: each guide must take at least one tourist. Count how many different groupings of guides and tourists are possible.

Pick an answer.

(A)
56
(B)
58
(C)
60
(D)
62
(E)
64

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

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.

1STEP 1

Count every split, rule ignored

Ignore the rule for a moment: each of the six tourists freely picks A or B, so the splits multiply to 2⁶ = 64.

2 × 2 × 2 × 2 × 2 × 2 = 2⁶ = 64
2STEP 2

Find the forbidden splits

A guide is left empty only when all six tourists pick the other guide, so exactly 2 of those splits break the rule.

forbidden = 1_all to A + 1_all to B = 2
3STEP 3

Subtract to get the answer

Subtract the forbidden pair from the total: 64 - 2 = 62 groupings leave both guides with at least one tourist — choice (D).

64 - 2 = 62 → (D)
Answer
62
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.
💡Key takeaway

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.

  • Count every split, rule ignored
  • Find the forbidden splits
  • Subtract to get the answer

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