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A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three

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A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 14 Jan 2010, 13:21
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A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three colors, red, blue, and green are used to paint the six faces of the cube. If the adjacent faces are painted with the different colors, in how many ways can the cube be painted?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 27
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Dec 2012, 02:53, edited 1 time in total.
Renamed the topic.
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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 20 Dec 2012, 02:59
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apoorvasrivastva wrote:
A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three colors, red, blue, and green are used to paint the six faces of the cube. If the adjacent faces are painted with the different colors, in how many ways can the cube be painted?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 27


If the base of the cube is red, then in order the adjacent faces to be painted with the different colors, the top must also be red. 4 side faces can be painted in Green-Blue-Green-Blue OR Blue-Green-Blue-Green (2 options).

But we can have the base painted in either of the three colors, thus the total number of ways to paint the cube is 3*2=6.

Answer: B.
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Re: Permutations Again :) [#permalink] New post 14 Jan 2010, 13:37
apoorvasrivastva wrote:
A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three colors, red, blue, and green are used to
paint the six faces of the cube. If the adjacent faces are painted with the different colors, in how many
ways can the cube be painted?
(A) 3
(B) 6
(C) 8
(D) 12
(E) 27

OA is
[Reveal] Spoiler:
B


Am I missing some thing here?

Any one side of the cube will be having 4 sides which can be termed as adjacent.. Isn't it ? Say for e.g side A it will be having 4 sides adjacent to it, one on left, one on right, one above and one below.

Please correct me if I am assuming this wrong..!

thanks
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Re: Permutations Again :) [#permalink] New post 14 Jan 2010, 13:59
nitishmahajan, I agree with your assessment. There are 4 adjacent sides for every face of the cube.

Let's say side 1 is painted red, then the 4 adjacent sides can be either green or blue alternating. This can be done in 2 ways.
GBGB
BGBG
Sixth side should be the same color as side 1.

For each color chosen for side 1(and side6) there are 2 ways of painting side 2,3,4 and 5.
No. of colors that can be chosen for side 1(and side6) is 3.
So 3*2 = 6..

Good question!!
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Re: Permutations Again :) [#permalink] New post 19 Dec 2012, 23:48
How do you solve this one?
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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 28 Dec 2012, 00:48
So, do they asked this on test day? This drove me nuts.

Imagine a flattened cube... The three colored region will establish the other colors of the remaining faces of the cube.

For example: We assumed the sequence of color in the given image as RED on face#1 and BLUE on face#2 and GREEN on face#3. Since face#1 is RED then we know #4 and #5 cannot be RED. Since face #2 is BLUE, we know that #5 and #6 cannot be BLUE. Since face#3 is GREEN, we know #4 and #6 (the bottom) cannot be GREEN.

So, all we need is to count the possible number of arrangements of 3 colors.

3! = 6
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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 28 Dec 2012, 18:30
mbaiseasy wrote:
So, do they asked this on test day? This drove me nuts.

Imagine a flattened cube... The three colored region will establish the other colors of the remaining faces of the cube.

For example: We assumed the sequence of color in the given image as RED on face#1 and BLUE on face#2 and GREEN on face#3. Since face#1 is RED then we know #4 and #5 cannot be RED. Since face #2 is BLUE, we know that #5 and #6 cannot be BLUE. Since face#3 is GREEN, we know #4 and #6 (the bottom) cannot be GREEN.

So, all we need is to count the possible number of arrangements of 3 colors.

3! = 6


Yes these are the questions that GMAT will ask I think its from the quant review or from gmat prep, they want you to apply the knowledge :)
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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 12 Sep 2013, 02:26
How about this approach?
There are 3 colors and 6 sides. Same color can't be next to each other so put them on opposite sides.
There are 3 opposite sides. So 3 colors 3 sides- no. of ways 3*2*1 = 6

Any flaw in this thinking?

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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three [#permalink] New post 13 Feb 2014, 08:00
Actually this one works out like this. Cube has 6 faces and we are told adjacent are different therefore base different from side different from front. Three sides to choose the colors to paint them with. Well since we have three colors then 3! =6 (B) is the right answer

Hope it clarifies
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Re: A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three   [#permalink] 13 Feb 2014, 08:00
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