AMC 8 · 2004 · #17

Grade 4 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

combinations-basicset-partitionsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: systematic-enumerationcombinations-basic

📏 Short solution 💡 2 insights

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Problem

Three friends share a total of 6 identical pencils, and each friend must have at least one pencil. How many different ways can the pencils be distributed among the three friends?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

With only 6 pencils and 3 friends, the unordered splits of 6 into three positive parts can be listed by hand. Tool #2 (Make a Systematic List) is the right primary — organize the splits from largest first to avoid missing any. Tool #9 (Solve an Easier Problem) trims the work: instead of listing every ordered triple, first find the unordered splits (an easier sub-problem) and then count arrangements of each split. Together they beat reaching for algebra on a problem this small.

1STEP 1

Write 6 as three positive parts, largest first, to find every unordered split: 4+1+1, 3+2+1, 2+2+2.

6 = 4+1+1 | 6 = 3+2+1 | 6 = 2+2+2

2STEP 2

The friends are distinct, so each split's arrangements are 3, 6, and 1 for 4+1+1, 3+2+1, 2+2+2.

split & repeats & arrangements ; 4+1+1 & two 1s & 3 ; 3+2+1 & all different & 6 ; 2+2+2 & all same & 1

3STEP 3

Add the arrangement counts: 3 + 6 + 1 = 10 ways, choice (D).

3 + 6 + 1 = 10 → (D)

Answer

Sanity check by listing the 10 ordered triples directly. From (4,1,1): (4,1,1), (1,4,1), (1,1,4) — that is 3. From (3,2,1): (3,2,1), (3,1,2), (2,3,1), (2,1,3), (1,3,2), (1,2,3) — that is 6. From (2,2,2): (2,2,2) — that is 1. Total 3 + 6 + 1 = 10, matching (D). Distractor sweep: (A) 1 would mean only (2,2,2) counts (forgets distinct friends), (B) 3 counts only (4,1,1) arrangements, (C) 6 counts only (3,2,1) arrangements, and (E) 12 over-counts by treating (4,1,1) as 6 arrangements instead of 3.

💡Key takeaway

When a problem says "how many ways," list the unordered splits first and then count seat arrangements for each — for 6 pencils among 3 friends, that gives 3 + 6 + 1 = 10.

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