Jevan must paint 3 rooms in a house. Room A can be painted

BTGmoderatorDC 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. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
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- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
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- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

BTGmoderatorDC 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?

1) 24
2) 26
3) 27
4) 63
5) 64

If there were no restrictions, there would be 3 x 3 x 3 = 27 ways to paint the 3 rooms. The only way they could be painted the same color would be if they were all painted red. Therefore, there are 27 - 1 = 26 ways to paint the 3 rooms such that all three rooms are not painted the same color.

Answer: B/26