2014 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2014 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:arrangements with restrictionspermutationscasework

Difficulty rating: 1140

4.

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

Solution:

Blue must come before yellow but not next to it, so they sit in positions (1,3)(1,3) or (1,4)(1,4) or (2,4)(2,4).

In each case orange and red fill the two remaining spots with orange before red, which is forced. The three orderings are orange, blue, red, yellow; blue, orange, red, yellow; and blue, orange, yellow, red.

There are 33 possible orderings.

Thus, B is the correct answer.

Problem 4 in Other Years