AMC 10 · 2024 · #20

Grade 7 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Arrangement Block Type
permutations-basiccombinations-basicsystematic-enumerationpattern-recognition systematic-enumerationcaseworkidentify-subproblems ↑ Prerequisites: permutations-basiccombinations-basicfactorial
📏 Medium solution 💡 3 insights
Problem
Three different pairs of shoes — call them P₁, P₂, P₃, each with a left shoe L_i and a right shoe R_i — are placed in a row of 6 positions. The rule: no left shoe is next to a right shoe from a different pair (i.e., an L_i R_j or R_j L_i adjacency is allowed only if i = j). How many orderings of the six distinct shoes obey the rule?

Pick an answer.

(A)
60
(B)
72
(C)
90
(D)
108
(E)
120

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The shoes look complicated, but the only thing the rule cares about is boundaries between L's and R's in the row. Tool #2 (Make a Systematic List) gives a clean two-layer plan: layer one — list every L/R type-string of three L's and three R's that has no forbidden boundary; layer two — for each surviving type-string, count how many ways to label the L's as L₁, L₂, L₃ and the R's as R₁, R₂, R₃ so each L-R boundary is a matched pair. Tool #7 (Identify Subproblems) splits the type-strings into four block-shape cases. Tool #9 (Solve an Easier Related Problem) reduces the labeling layer to a small permutation count per case.

1STEP 1

Read only the L/R skeleton: every L-R boundary must be a matched pair, so no LRL or RLR can appear as a type-substring.

boundary L_i R_j → i = j; LRL, RLR forbidden as type-substrings
2STEP 2

List the type-strings of three L's and three R's with no LRL or RLR — exactly 8 survive, in three groups by boundary count.

valid type-strings = {LLLRRR, RRRLLL, LLRRRL, LRRRLL, RLLLRR, RRLLLR, LRRLLR, RLLRRL}
3STEP 3

Group A (1 boundary: LLLRRR, RRRLLL): fix the boundary pair (3 ways), order the free shoes each side — 12 per string, subtotal 24.

3 · 2! · 2! = 12 each, subtotal = 12 + 12 = 24
4STEP 4

Group B (2 boundaries: 4 strings): pick an ordered pair of distinct matched pairs, P(3,2)=6, rest forced — 6 per string, subtotal 24.

P(3, 2) = 6 each, subtotal = 4 · 6 = 24
5STEP 5

Group C (3 boundaries: LRRLLR, RLLRRL): the three boundaries use the three pairs in some order, 3!=6 per string, subtotal 12.

3! = 6 each, subtotal = 2 · 6 = 12
6STEP 6

Add the three disjoint subtotals — 24 + 24 + 12 gives 60, choice (A).

Total = 24 + 24 + 12 = 60 → (A)
Answer
60
Cross-check the totals by an independent route. Each of the 8 valid type-strings was placed in one of three classes — 1, 2, or 3 boundaries — and the per-string counts 12, 6, 6 depend only on the number of boundaries: 12 = 3 · (2!)² for one boundary, 6 = 3 · 2 for two boundaries (ordered pair of distinct pairs), 6 = 3! for three boundaries. Aggregating: 2 · 12 + 4 · 6 + 2 · 6 = 24 + 24 + 12 = 60. The total 60 is well below the unconstrained 6! = 720, consistent with a real adjacency restriction. Answer (A) = 60.
💡Key takeaway

This AMC 10 problem only needs Grade 7 organized counting — list every valid L/R skeleton, then count name choices at each boundary — and the answer drops out as 60!

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