AMC 10 · 2025 · #18

Grade 11 counting

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

combinations-basiccomplementary-countingsymmetry-argumentfactorial complementary-countingsystematic-enumeration ↑ Prerequisites: combinations-basic

📏 Long solution 💡 3 insights

Problem

Count the ordered triples (x, y, z) built from three different positive integers, each from 1 to 8, for which the product of every two of them is larger than the remaining one: xy ＞ z, xz ＞ y, and yz ＞ x all hold at once.

Pick an answer.

(A)

(B)

(C)

186

(D)

336

(E)

486

How to solve

Strategy Change Focus / Count the Complement

The three conditions treat x, y, z identically, so Tool #15 (Organize Information in More Ways) lets us stop tracking order and study only the underlying set {x, y, z} with x ＜ y ＜ z, restoring order at the end with a × 6 factor. Tool #14 (Extreme Principle) then trains attention on the smallest value: it rules out 1 instantly and shows that once the smallest number is at least 2, two of the three inequalities can never fail — collapsing the whole problem to the single condition xy ＞ z. Counting the sets that satisfy xy ＞ z head-on is awkward, so Tool #16 (Count the Complement) flips it: count all 3-element sets and subtract the few that fail, and Tool #2 (Make a Systematic List) produces that short failure list by hand.

1STEP 1

Drop the order, keep the set

All three rules treat the numbers alike, so count unordered sets with x ＜ y ＜ z, then restore order by multiplying by 3! = 6.

(ordered valid triples) = 6 × (valid sets with x ＜ y ＜ z)

2STEP 2

The number 1 is impossible

If the smallest value is 1, then xy ＞ z and xz ＞ y demand y ＞ z and z ＞ y — impossible. So every value is at least 2, from {2,…,8}.

x = 1 → y ＞ z and z ＞ y (contradiction) → x ≥ 2

3STEP 3

Only one inequality can fail

With x ＜ y ＜ z and x ≥ 2, both yz ＞ x and xz ＞ y hold automatically, so only xy ＞ z can ever fail.

yz ≥ y ＞ x, xz ≥ 2z ＞ 2y ＞ y → only xy ＞ z matters

4STEP 4

Count all sets, plan to subtract

Count every 3-element set of {2,…,8} — there are C(7, 3) = 35 — then subtract those that break xy ＞ z.

C(7, 3) = 35, (valid sets) = 35 - (sets with xy ≤ z)

5STEP 5

List the failures by hand

A failing set needs xy ≤ z ≤ 8: (2,3) allows z ∈ {6,7,8}, (2,4) allows z = 8, larger pairs exceed 8 — 4 failures.

(2,3){:} z ∈ {6,7,8}; (2,4){:} z = 8; all other pairs give xy ＞ 8 → 4 failures

6STEP 6

Assemble the count

Valid sets number 35 - 4 = 31, and each fans out into 6 ordered triples: 31 × 6 = 186, which is choice (C).

(35 - 4) × 6 = 31 × 6 = 186 → (C)

Answer

186

The count sits sensibly below the ceiling: three different values from 1–8 form 8 · 7 · 6 = 336 ordered triples with no conditions at all, so the answer cannot exceed 336 — that alone eliminates (D) 336 and (E) 486. Peeling off the impossible cases confirms 186: of the C(8, 3) = 56 sets, the C(7, 2) = 21 that contain a 1 and the 4 that fail xy ＞ z are all discarded, leaving 56 - 21 - 4 = 31 sets, or 31 × 6 = 186 ordered triples. Spot check: the set {2, 3, 6} fails because 2 · 3 = 6 is not greater than 6, while {2, 3, 4} works since 2 · 3 = 6 ＞ 4, 2 · 4 = 8 ＞ 3, and 3 · 4 = 12 ＞ 2.

💡Key takeaway

Because the rules do not care about order, count the three-number sets instead, notice that only the two smallest numbers can ever break the rule, subtract those few failures, then multiply by 6.

Drop the order, keep the set
The number 1 is impossible
Only one inequality can fail
Count all sets, plan to subtract
List the failures by hand
Assemble the count