AMC 10 · 2018 · #4

Grade 7 counting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

combinations-basicpermutations-basicsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: combinations-basicpermutations-basic

📏 Medium solution 💡 2 insights

Problem

A student has a 6-period day and must schedule 3 different mathematics courses — algebra, geometry, and number theory — one course per period. No two of these math courses may sit in consecutive periods. The other 3 periods can be anything and do not matter. The question asks how many different schedules of the three math courses are possible.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The phrase "how many ways" points to Tool #2 (Make a Systematic List): when the no-two-consecutive rule is tricky, the safest move is to list every valid set of periods in order so none are missed and none are double-counted. Tool #7 (Identify Subproblems) splits the question into two independent pieces — first "which three periods?" (where the courses go) and then "in what order?" (which course goes in each slot) — so the two small counts can be multiplied. Tool #1 (Draw a Diagram) backs up the listing by picturing the periods as six slots, making it easy to see when two math classes touch.

1STEP 1

Split into where, then which order

Two independent choices: which 3 periods hold math (no two consecutive), and which course fills each. Count each, then multiply.

(ways to pick the periods) × (ways to order the 3 courses)

2STEP 2

List the valid period sets

List every triple of periods 1–6 with no two adjacent: {1,3,5}, {1,3,6}, {1,4,6}, {2,4,6} — 4 valid sets in all.

{1,3,5}, {1,3,6}, {1,4,6}, {2,4,6} → 4

3STEP 3

Count the orders of three courses

With the three periods fixed, three distinct courses fill them: 3 × 2 × 1 = 6 orders.

3 × 2 × 1 = 6

4STEP 4

Multiply the two counts

Each of the 4 sets fills 6 ways with none repeating: 4 × 6 = 24 schedules, matching choice (E).

4 × 6 = 24 → (E)

Answer

The two pieces are believable on their own: there really are only 4 ways to spread three periods apart in a row of six, and arranging three distinct courses always gives 3! = 6 orders, so 4 × 6 = 24 is consistent. Note that 24 must be a multiple of 6 because every period set is counted with all 6 orderings — and among the choices only 6, 12, 18, 24 are multiples of 6, while the four-set count 4 then forces exactly 24. The smaller answers like (A) 3 or (B) 6 would mean almost no freedom in placement, which contradicts the four clearly different period sets we listed.

💡Key takeaway

Count where the three classes can go (4 spread-out slot sets) and how the three different classes can be ordered (6 ways), then multiply: 4 times 6 is 24.

Split into where, then which order
List the valid period sets
Count the orders of three courses
Multiply the two counts

Split into where, then which order

List the valid period sets

Count the orders of three courses

Multiply the two counts

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