AMC 10 · 2014 · #4

Grade 2 logiccounting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

permutations-basiclogical-deduction systematic-enumerationcasework ↑ Prerequisites: permutations-basic

📏 Medium solution 💡 2 insights

Problem

Four houses stand in a row, one orange, one red, one blue, and one yellow. The orange house comes somewhere before the red house, the blue house comes somewhere before the yellow house, and the blue and yellow houses are not right next to each other. Count how many left-to-right orderings of the four colors obey all three rules.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only 4!=24 orderings, a small finite world, so Tool #2 (Make a Systematic List) can sweep it without missing or repeating a case. To make the rules mechanical, Tool #4 (Introduce a Variable) numbers the four spots 1,2,3,4, turning "before" into "smaller spot number" and "not next to each other" into "spot numbers differ by at least 2." The trick that keeps the list short is to place blue and yellow first, because those two colors carry two of the three rules; Tool #3 (Eliminate Possibilities) then throws out every blue-yellow placement that is adjacent, leaving only a handful to finish.

1STEP 1

Number the four spots

Number the spots 1,2,3,4 from the left: "before" becomes a smaller spot number, "not next to each other" becomes a gap of at least 2.

spots 1,2,3,4; O ＜ R, B ＜ Y, |B-Y| ≥ 2

2STEP 2

Place blue and yellow first

Blue and yellow carry two rules, so place them first: of the six spot pairs, the touching ones (1,2),(2,3),(3,4) go, leaving three.

keep (B,Y)∈{(1,3),(1,4),(2,4)}; drop (1,2),(2,3),(3,4)

3STEP 3

Fill in orange and red

The two leftover spots must go orange then red, so (1,3) gives BOYR, (1,4) gives BORY, and (2,4) gives OBRY.

(1,3)→ BOYR, (1,4)→ BORY, (2,4)→ OBRY

4STEP 4

Count the finished orderings

Each legal blue-yellow placement finished in exactly one way, so 3 orderings survive — choice (B).

3 valid orderings → (B)

Answer

Cross-check the three survivors by hand: BOYR (blue spot 1, yellow spot 3, gap 2), BORY (blue 1, yellow 4, gap 3), OBRY (blue 2, yellow 4, gap 2) — all keep blue before yellow with a gap of at least 2, and in each one orange sits before red. So 3 is achievable, ruling out (A) 2 as too small. It also cannot be as large as 6: even before the not-adjacent rule, only 6 of the 24 orderings have both orange-before-red and blue-before-yellow, and the adjacency rule then removes some of those, so (E) 6 is impossible.

💡Key takeaway

Number the four spots, place blue and yellow first with a gap between them, and "orange before red" fills in the rest — only three rows survive.

Number the four spots
Place blue and yellow first
Fill in orange and red
Count the finished orderings

Number the four spots

Place blue and yellow first

Fill in orange and red

Count the finished orderings

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