AJMO Counting: Organise Cases Without Repeats or Missing Possibilities

By Ace Achievers · Published 16 October 2026 · Official information checked 14 October 2026

What are we counting?

The official samples include adjacency and distinct-item restrictions, as well as sums where changing the order does not create a new outcome. Read that distinction before calculating. An ordered route A–B–E differs from E–B–A; an unordered collection of those three labels does not.

Our original examples make the rules explicit. Before listing cases, state whether positions are fixed, whether an item can be reused, and whether a rotation or reversal changes the outcome. A correct arithmetic total built on the wrong definition is still the wrong count.

How can fixing the middle position simplify a route?

Count ordered routes using three different tiles, where each move crosses a shared edge. If the middle tile has d neighbours, there are d choices for the first tile and d−1 choices for the last: the first and last must differ. Thus that middle position contributes d(d−1) routes.

Every valid route has exactly one middle tile, so the cases cannot overlap. Each permitted pair of different neighbours gives one route, so the count also misses none. This completeness explanation matters as much as the calculation. As a check, reversing a route gives a different valid route, and twenty is even.

Change the board to a 2×2 square. Each of four tiles has two neighbours, giving 4×2×1 = 8 ordered routes. Adding diagonal moves changes the neighbour counts, so the earlier answer cannot simply be reused.

How do non-crossing pairings create smaller cases?

Six labelled positions are fixed around a circle. Join them into three pairs with straight chords that do not cross. Position 1 can be paired with 2, 4 or 6. A partner in an odd-labelled position would leave an odd number of points on each side; those points cannot pair internally without a crossing.

Pairing 1 with 2 leaves four points, which have two non-crossing pairings. Pairing 1 with 4 leaves two points on each side, giving one possibility. Pairing 1 with 6 again leaves four points, giving two. The total is 2 + 1 + 2 = 5. The positions are fixed, so we do not also count different seating arrangements.

The four-point count is easy to check directly: two adjacent-pair patterns work, while pairing opposite points crosses. For eight fixed points, the same reasoning gives 5 + 2 + 2 + 5 = 14. Derive this from the smaller counts rather than starting with a named formula.

How do we test a counting argument?

Test a small instance, label the cases, and check why every outcome enters exactly one case. If order is irrelevant, a condition such as a<b<c can avoid listing permutations of the same collection. If order matters, do not divide by a symmetry merely because the picture looks similar.

Keep a practice note stating the counting decision, not only the final number. That note makes the method reusable when the next question changes the board, the number of positions or the repetition rule.