AMC 10 · 2014 · #4
Easy mode Grade 2Ralph walks past four houses in a row. Each house is a different color: orange, red, blue, and yellow. He passes the orange house before the red house. He passes the blue house before the yellow house. The blue house and the yellow house are not next to each other. In how many orders can the four colored houses be arranged?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Four houses in a row, all different colors: orange ($O$), red ($R$), blue ($B$), yellow ($Y$); Orange is somewhere before red ($O$ earlier in the row than $R$); Blue is somewhere before yellow ($B$ earlier in the row than $Y$); Blue and yellow are not next to each other (at least one house sits between them); Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6
Unknowns: The number of color orderings of the four houses that satisfy all three rules
Understand
Restated: 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.
Givens: Four houses in a row, all different colors: orange ($O$), red ($R$), blue ($B$), yellow ($Y$); Orange is somewhere before red ($O$ earlier in the row than $R$); Blue is somewhere before yellow ($B$ earlier in the row than $Y$); Blue and yellow are not next to each other (at least one house sits between them); Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
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.
Execute — Answer: B
2.OA.A.1 Step 1 Number the four spots
- Label the row of houses spot $1$, $2$, $3$, $4$ from left to right.
- Now each rule becomes a rule about spot numbers: "orange before red" means orange's spot number is smaller than red's; "blue before yellow" means blue's spot number is smaller than yellow's; "blue and yellow not next to each other" means their spot numbers differ by at least $2$.
💡 Numbering the spots turns "before" and "next to" into plain comparisons of whole numbers.
2.OA.B.2 Step 2 Place blue and yellow first
- Blue and yellow carry two of the three rules, so pin them down first.
- Choose which two spots hold blue and yellow, with blue in the smaller-numbered spot.
- The six possible spot pairs are $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$.
- Toss out the ones where blue and yellow touch — $(1,2)$, $(2,3)$, and $(3,4)$ — because their spot numbers differ by only $1$.
- That leaves exactly three legal blue-yellow placements.
💡 Starting with the two colors that share the most rules kills the illegal cases right away.
2.OA.A.1 Step 3 Fill in orange and red
- For each surviving blue-yellow placement, the two leftover spots must hold orange and red.
- The rule "orange before red" forces orange into the smaller leftover spot and red into the larger one, so there is exactly one way to finish each case.
- Placement $(B,Y)=(1,3)$ leaves spots $2,4$, giving $B\,O\,Y\,R$.
- Placement $(1,4)$ leaves spots $2,3$, giving $B\,O\,R\,Y$.
- Placement $(2,4)$ leaves spots $1,3$, giving $O\,B\,R\,Y$.
💡 Once blue and yellow sit down, "orange before red" leaves no choice for the last two houses.
2.OA.A.1 Step 4 Count the finished orderings
- Each of the three legal blue-yellow placements produced exactly one full ordering, and all three — $B\,O\,Y\,R$, $B\,O\,R\,Y$, and $O\,B\,R\,Y$ — pass every rule: orange precedes red, blue precedes yellow, and blue and yellow never touch.
- No other placements survived, so the total number of valid orderings is $3$, which is choice (B).
💡 Three legal blue-yellow spots, each finishing in just one way, add up to three orderings.
2.OA.A.1 Label the row of houses spot $1$, $2$, $3$, $4$ from left to right. Now each rul 2.OA.B.2 Blue and yellow carry two of the three rules, so pin them down first. Choose whi 2.OA.A.1 For each surviving blue-yellow placement, the two leftover spots must hold orang 2.OA.A.1 Each of the three legal blue-yellow placements produced exactly one full orderin Review
Reasonableness: 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.
Alternative: Count by subtraction. Of the $24$ orderings, exactly half have orange before red, and half of those have blue before yellow, giving $24\div2\div2=6$ orderings that respect both "before" rules. Among those $6$, the bad ones glue blue and yellow into a single "$BY$" block; that block together with orange and red is three items, and "orange before red" leaves $3$ arrangements, all bad. Subtracting, $6-3=3$, the same answer (B).
CCSS standards used (min grade 2)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Translating the "before" and "not next to" wording into rules about spot numbers, then counting the finished orderings that obey them.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Finding the gap between blue's and yellow's spot numbers to reject any placement where they differ by only 1.)
⭐ 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 with a gap between them, and "orange before red" fills in the rest — only three rows survive.
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