AMC 8 · 2015 · #4
Easy mode Grade 4Problem
The Blue Bird High School chess team has 5 people: 2 boys and 3 girls.
A photographer wants them to sit in one row for a team photo. She has one rule: a boy sits at the left end, a boy sits at the right end, and the 3 girls sit in the middle three seats.
How many different seating arrangements are possible?
Pick an answer.
AMC 8 2015 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: A chess team has $2$ boys and $3$ girls. They line up for a photo with a boy at the left end, a boy at the right end, and the three girls filling the three middle spots. How many different lineups are possible?
Givens: Team members: $2$ boys (distinct people) and $3$ girls (distinct people); Seat layout (left to right): boy, girl, girl, girl, boy; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $6$, (E) $12$
Unknowns: The total number of distinct left-to-right arrangements of the $5$ team members under the seat layout
Understand
Restated: A chess team has $2$ boys and $3$ girls. They line up for a photo with a boy at the left end, a boy at the right end, and the three girls filling the three middle spots. How many different lineups are possible?
Givens: Team members: $2$ boys (distinct people) and $3$ girls (distinct people); Seat layout (left to right): boy, girl, girl, girl, boy; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $6$, (E) $12$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
The seat layout splits cleanly into two independent jobs: (a) place the $2$ boys in the $2$ end seats, and (b) place the $3$ girls in the $3$ middle seats. Tool #7 (Identify Subproblems) handles this split — once we count each piece, the multiplication principle combines them. Tool #2 (Systematic List) is the safety net for counting each piece: list boy orderings (just $2$) and girl orderings (only $6$) in a fixed order so nothing is missed or doubled.
Execute — Answer: E
3.OA.A.1 Step 1 - Count the boy arrangements at the two ends.
- Call the boys $B_1$ and $B_2$.
- Either $B_1$ sits on the left and $B_2$ on the right, or the reverse.
- That is $2$ choices for the left end, and once chosen, only $1$ boy remains for the right end.
💡 "Choices for the first seat times choices for the second" is Grade 3 multiplication as combining groups of options.
3.OA.A.1 Step 2 - Count the girl arrangements in the three middle seats.
- Call the girls $G_1, G_2, G_3$.
- The leftmost middle seat has $3$ girls to choose from; after one sits, the next seat has $2$ choices left; the last seat has $1$ girl forced.
- List them in order to double-check: $G_1G_2G_3$, $G_1G_3G_2$, $G_2G_1G_3$, $G_2G_3G_1$, $G_3G_1G_2$, $G_3G_2G_1$ — exactly $6$.
💡 Listing all $6$ orderings in a fixed alphabetical rule is the Tool #2 systematic-list move that confirms the multiplication.
4.OA.A.3 Step 3 - Combine the two independent subproblems.
- Every boy arrangement can pair with every girl arrangement, so multiply the two counts.
💡 Solving a multi-step word problem by counting each part and multiplying the results is the Grade 4 multi-step operations standard.
3.OA.A.1 Count the boy arrangements at the two ends. Call the boys $B_1$ and $B_2$. Eithe 3.OA.A.1 Count the girl arrangements in the three middle seats. Call the girls $G_1, G_2, 4.OA.A.3 Combine the two independent subproblems. Every boy arrangement can pair with eve Review
Reasonableness: Sanity check the size. Without any seat restriction, $5$ people in a row give $5! = 120$ lineups. Forcing the two boys to the ends keeps only the fraction where the end seats happen to hold the boys. The chance that any one given end seat is filled by a boy when seating randomly is $\tfrac{2}{5}$, and then the other end being the remaining boy is $\tfrac{1}{4}$, so the fraction is $\tfrac{2}{5} \times \tfrac{1}{4} = \tfrac{1}{10}$. Indeed $120 \times \tfrac{1}{10} = 12$, matching answer (E).
Alternative: Tool #3 (Eliminate Possibilities) on the choices: the boys contribute a factor of $2$ (since $2! = 2$) and the girls contribute a factor of $6$ (since $3! = 6$), so the total must be a multiple of both $2$ and $6$. Only (D) $6$ and (E) $12$ are multiples of $6$, and (D) ignores the boy factor — so (E) $12$ is forced.
CCSS standards used (min grade 4)
3.OA.A.1Interpret products of whole numbers as combining groups of equal size (Counting boy arrangements as $2 \times 1 = 2$ and girl arrangements as $3 \times 2 \times 1 = 6$ by multiplying the choices available at each seat.)4.OA.A.3Solve multistep word problems using the four operations (Combining the two subproblem counts (boys: $2$, girls: $6$) by the multiplication step $2 \times 6 = 12$ to get the total number of lineups.)
⭐ This AMC 8 problem only needs Grade 4 multi-step multiplication — count each part, then multiply!
⭐ This AMC 8 problem only needs Grade 4 multi-step multiplication — count each part, then multiply!
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