2014 AMC 12A Problem 3

Attempt Problem 3 of the 2014 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
22
33
44
55
66
Answer: B
Concepts:arrangements with restrictionscasework
Small hint:
Split into cases by whether the orange house or the blue house comes first
Big hint:
Because blue and yellow cannot be adjacent, at least one house must separate them
Solution:
If orange comes first, then blue and yellow cannot be adjacent, forcing the order orange, blue, red, yellow.
If blue comes first, yellow can be in the third or fourth position (never second, to avoid adjacency), giving blue, orange, yellow, red and blue, orange, red, yellow.
These are the only 33 valid orderings.
Thus, the correct answer is B.
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