AMC 8 · 2004 · #17
Grade 4 countingProblem
Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen?
Pick an answer.
AMC 8 2004 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: 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?
Givens: There are $6$ identical pencils in total; There are $3$ distinct friends; Each friend gets at least $1$ pencil; Answer choices: (A) $1$, (B) $3$, (C) $6$, (D) $10$, (E) $12$
Unknowns: The number of distinct distributions $(x_1, x_2, x_3)$ with $x_1 + x_2 + x_3 = 6$ and each $x_i \ge 1$
Understand
Restated: 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?
Givens: There are $6$ identical pencils in total; There are $3$ distinct friends; Each friend gets at least $1$ pencil; Answer choices: (A) $1$, (B) $3$, (C) $6$, (D) $10$, (E) $12$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #9 Solve an Easier Problem
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.
Execute — Answer: D
4.OA.B.4 Step 1 - List the unordered splits of $6$ into three positive parts.
- Write each split with the largest part first so duplicates are easy to spot.
- Start with the biggest possible top part and walk down.
💡 Grade 4 factor/partition thinking: writing parts in non-increasing order is the standard way to enumerate splits without repeats. Only three splits exist for $6$ into three positive parts.
4.OA.A.3 Step 2 - Count seat arrangements for each split.
- The three friends are distinct, so for each split we count how many ways the three numbers can be assigned to the three friends.
- Use a small chart.
💡 Place the unique number in any of $3$ seats for $(4,1,1)$ — $3$ ways. For $(3,2,1)$ every order is different, giving $3 \times 2 \times 1 = 6$ ways. For $(2,2,2)$ everyone gets the same count, so there is only $1$ way.
4.OA.A.3 Step 3 Add the arrangement counts to get the total number of ways.
💡 Each split's arrangements are different distributions (different friend gets the bigger pile), so summing covers every case exactly once.
4.OA.B.4 List the unordered splits of $6$ into three positive parts. Write each split wit 4.OA.A.3 Count seat arrangements for each split. The three friends are distinct, so for e 4.OA.A.3 Add the arrangement counts to get the total number of ways. Review
Reasonableness: 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 $3$.
Alternative: Tool #9 (Solve an Easier Problem) by pre-assignment: give each friend one pencil first to honor the "at least one" rule. That uses up $3$ pencils, leaving $3$ pencils to share freely (zeros now allowed). The harder original problem becomes the easier problem of splitting $3$ identical pencils among $3$ friends with no minimum. Listing those: $(3,0,0)$ has $3$ arrangements, $(2,1,0)$ has $6$, $(1,1,1)$ has $1$, total $3+6+1=10$, again (D).
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using the four operations (Counting the arrangements of each split (using multiplication and addition) and adding the case counts $3 + 6 + 1 = 10$.)4.OA.B.4Find all factor pairs for a whole number in the range 1-100; recognize that a whole number is a multiple of each of its factors (Enumerating the unordered splits of $6$ into three positive parts in non-increasing order — the same systematic enumeration habit students use to find factor pairs.)
⭐ 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 $3 + 6 + 1 = 10$.
⭐ 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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