Princ wrote:
Jevan must paint 3 rooms in a house.Room A can be painted orange,red or green.Room B can be painted orange,white or red.Room C can be painted white,red or green.The 3 rooms cannot all be painted the same color.In how many different ways could Jevan paint the 3 rooms?
A) 24
B) 26
C) 27
D) 63
E) 64
Take the task of painting the 3 rooms and break it into
stages.
Stage 1: Select a color for Room A
Room A can be painted orange, red or green
So, we can complete stage 1 in
3 ways
Stage 2: Select a color for Room B
Room B can be painted orange, white or red.
So we can complete this stage in
3 ways.
Stage 3: Select a color for Room C
Room C can be painted white, red or green.
So we can complete this stage in
3 ways.
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus paint all 3 rooms) in
(3)(3)(3) ways (= 27 ways)
However, some of these 27 possible outcomes BREAK the condition that
the 3 rooms cannot all be painted the same color.Given the different color options for each room, we can see that the ONLY way that the 3 rooms can be the same color is when all 3 rooms are painted RED.
Since there is only 1 way to BREAK the given condition, the number of GOOD outcomes = 27 - 1 = 26
Answer: B
Note: the FCP can be used to solve the
MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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