AMC 10 · 2014 · #3

Grade 2 logiccounting
permutations-basiclogical-deduction systematic-enumerationcasework ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 2 insights
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Problem
Two ordering rules apply and one pair must not be adjacent. Count the arrangements.

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
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

Numbering the spots makes the rules concrete.

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

Place blue and yellow first

The tighter pair has only three placements.

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

Fill in orange and red

The other two are then forced.

(1,3)→ BOYR, (1,4)→ BORY, (2,4)→ OBRY
4STEP 4

Count the finished orderings

So there are 3 orderings, choice (B).

3 valid orderings → (B)
Answer
3
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