AMC 8 · 2015 · #4

Grade 4 counting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

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

📏 Short solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

The left end offers 2 boys to choose and the right end is then forced to the last boy, giving 2 boy arrangements.

2 × 1 = 2 boy arrangements

2STEP 2

The three middle seats take 3 girls, then 2, then 1 — listing every order confirms exactly 6 girl arrangements.

3 × 2 × 1 = 6 girl arrangements

3STEP 3

Every boy arrangement pairs with every girl arrangement, so multiply the two counts: 2 × 6.

total = 2 × 6 = 12 → (E)

Answer

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

\frac{2}{5}

, and then the other end being the remaining boy is

\frac{1}{4}

, so the fraction is

\frac{2}{5}

\frac{1}{4}

\frac{1}{10}

. Indeed 120 ×

\frac{1}{10}

= 12, matching answer (E).

💡Key takeaway

This AMC 8 problem only needs Grade 4 multi-step multiplication — count each part, then multiply!

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