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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pm; Followed by:23 members

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

by BTGmoderatorDC » Sun Sep 30, 2018 6:06 pm

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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

OA B

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Sun Sep 30, 2018 6:06 pm

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Mon Oct 01, 2018 4:50 am

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

OA B

Source: Magoosh

Without restriction, since each room can be painted in 3 ways, the total number of ways to paint the rooms is 3^3.

However, this includes those in which all the rooms are painted the same color. There is 1 way, all rooms red, that violate the restriction, so this must be subtracted, [spoiler]27-1=26, B[/spoiler]

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

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

by Brent@GMATPrepNow » Mon Oct 01, 2018 8:35 am

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
--------------------------

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Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Wed Oct 03, 2018 4:47 pm

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

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