AMC 10 · 2018 · #3
Grade 7 countingPick an answer.
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.
Split into where, then which order
Split into choosing slots and choosing an order.
Choosing the spots and choosing the order are independent, so handle them one at a time and combine.
Choosing the spots and choosing the order are independent, so they can be handled one at a time.
▸ Why?
Every choice of spots can be paired with every ordering, so the two counts multiply.
▸ Why?
Each finished schedule is built from exactly one pair of those choices, so nothing is counted twice.
List the valid period sets
Only four sets of periods work.
Listing the sets in increasing order guarantees you catch every spread-out triple exactly once.
7.SP.C.8Make A Systematic ListCount the orders of three courses
The three courses order in six ways.
Filling distinct slots with distinct courses is a shrinking choice: 3, then 2, then 1.
7.SP.C.8Make A Systematic ListMultiply the two counts
Multiplying gives 24.
Four placements, each opening up six orderings, multiply to the grand total.
3.OA.A.1Identify SubproblemsCount 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