AMC 8 · 2004 · #17
Easy mode Grade 4Problem
Three friends share pencils. The pencils all look the same.
Each friend must end up with at least one pencil. (So no one gets zero pencils.)
In how many different ways can the pencils be split between the three friends?
Pick an answer.
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 $3 + 6 + 1 = 10$.