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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?
Re: Jevan must paint 3 rooms in a house.Room can be painted orange,red or
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02 Jun 2018, 06:16
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Room A - Red, Orange, Green Room B - Red, Orange, White Room C - Red, Green, White
Total possible combinations = 3*3*3 = 27 Undesired combination = All rooms painted same color, here in our case only one possibility, Red color = 1 Desired possible combinations = 27-1 = 26 Hence, imo option 2
Thanks and regards, Sharan Salem _________________
Re: Jevan must paint 3 rooms in a house.Room can be painted orange,red or
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02 Jun 2018, 06:20
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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?
1) 24 2) 26 3) 27 4) 63 5) 64
Room A - Orange, Red or Green Room B - Orange, White or Red Room C - White, Red or Green
# ways to paint Room A = 3 ways # ways to paint Room B = 3 ways # ways to paint Room C = 3 ways
Total # ways the three rooms can be painted without any conditions = 3*3*3 = 27 ways
Now the condition that the three rooms will have the same color, is when all three are painted Red. It can be done in only 1 way.
Hence the Total # ways the 3 rooms can be painted = 27-1 = 26 ways.
Re: Jevan must paint 3 rooms in a house.Room can be painted orange,red or
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01 Oct 2018, 08:44
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Top Contributor
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.
Re: Jevan must paint 3 rooms in a house.Room can be painted orange,red or
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31 Aug 2021, 23:45
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