Competition · AMC preparation · step 4 of 4

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

List the splits of 6

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

Count arrangements for each split

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

Place the unique number in any of 3 seats for (4,1,1) — 3 ways. For (3,2,1) every order is different, giving 3 × 2 × 1 = 6 ways. For (2,2,2) everyone gets the same count, so there is only 1 way.

Handing a split's parts to the three distinct friends gives three different distributions for 4+1+1, six for 3+2+1, and one for 2+2+2.

▸ Why?

In 4+1+1 the two friends who each hold one pencil are interchangeable, so a distribution is pinned down entirely by which one of the three friends holds the pile of four — that is three choices.

▸ Why?

Each distinct distribution matches exactly one friend chosen to hold the larger pile, and relabeling the two friends who each hold one pencil produces the same counts, so the distributions pair one-for-one with the three friends.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

▸ Why?

In 3+2+1 all three parts differ, so any change in who holds which pile changes someone's count; the number of distributions is the number of ways to line the three distinct amounts up with the three friends.

▸ Why?

Assign the piles by two choices in turn — any of the three friends can take the pile of three, and then, whoever that was, either of the two friends who remain can take the pile of two while the last friend is forced to take the single — the second choice always has two options no matter how the first went, so the independent choices multiply, 3 × 2 = 6.

Independent choices multiplyIf a first choice can go a ways and each of those leaves b ways for a second, the pair can be made a times b ways.

▸ Why?

In 2+2+2 every friend holds the same two pencils, so no way of naming who-gets-what changes the list of counts, leaving exactly one distribution.

▸ Why?

Because all three counts are equal, every way of naming who holds what pairs with the single count-list where each friend holds two, so there is just one distinct distribution.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

4.OA.A.3Make A Systematic List

3STEP 3

Add the counts

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.

List the splits of 6
Count arrangements for each split
Add the counts

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List the splits of 6

Count arrangements for each split

Add the counts

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