AMC 8 · 2015 · #4

Grade 4 counting

permutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?

$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#2 Make a Systematic List 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.

$$2 \times 1 = 2 \text{ boy arrangements}$$

💡 "Choices for the first seat times choices for the second" is Grade 3 multiplication as combining groups of options.

#2 Make a Systematic List 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 $6$.

$$3 \times 2 \times 1 = 6 \text{ girl arrangements}$$

💡 Listing all $6$ orderings in a fixed alphabetical rule is the Tool #2 systematic-list move that confirms the multiplication.

#7 Identify Subproblems 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.

$$\text{total} = 2 \times 6 = 12 \;\Rightarrow\; \textbf{(E)}$$

💡 Solving a multi-step word problem by counting each part and multiplying the results is the Grade 4 multi-step operations standard.

[1] #2 3.OA.A.1 Count the boy arrangements at the two ends. Call the boys $B_1$ and $B_2$. Eithe

[2] #2 3.OA.A.1 Count the girl arrangements in the three middle seats. Call the girls $G_1, G_2,

[3] #7 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.1 Interpret 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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