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six-sided cube and six cans of paint

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by abhi332 » Thu Feb 25, 2010 12:51 pm
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How
many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to
look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

[spoiler]OA:B[/spoiler]
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by harsh.champ » Thu Feb 25, 2010 2:24 pm
abhi332 wrote:You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How
many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to
look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

[spoiler]OA:B[/spoiler]
Any 2 adjacent colors should be different in each case.
Hence we have 6 x 5 =30ways
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by kstv » Thu Feb 25, 2010 10:55 pm
The cube will have a different arrangement of colours only when the colours on the sides opposite to each other vary. The different ways of painting the opposite sides are 6 X 5 = 30.
Last edited by kstv on Wed Mar 10, 2010 8:28 am, edited 1 time in total.
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by stufigol » Thu Mar 04, 2010 4:39 pm
where did u find these pblem...plz
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by florencejennifer » Tue Mar 09, 2010 3:31 am
Six sides can be painted six colors in 6! Ways. But each number in the reorientation can take any of the six sides, so we have repetitions in the way of appearance.

i.e., the total number of such arrangements are 6!/6 = 120.

(This is almost similar to circular permutation.)

Correct Answer: E
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by Shawshank » Tue Mar 09, 2010 4:18 am
hi Florence,

Six sides can be painted six colors in 6! Ways. But each number in the reorientation can take any of the six sides, so we have repetitions in the way of appearance.

AM not clear what the bold face means,,, CAn you explain,,,
i.e., the total number of such arrangements are 6!/6 = 120.

(This is almost similar to circular permutation.)

Correct Answer: E
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by thebigkats » Fri Feb 11, 2011 8:22 pm
Hello:
I am unable to understand either of the solutions provided so far. Allow me to explain what solution I am reaching:

Lets assume the cube has sides Top, Bottom, North, East, South, West (TBNESW) and colors are abcdef)

North can take any of the 6 colors, for each of these colors, south can take any of the other 5 colors. So NS can be 6*5 = 30 combos

The other 4 colors on sides (NESW) can be arranged in 4! ways (= 4*3*2 = 24). However for each combination (say colors) abcd, there are 3 other combinations (bcda, cdab, dabc), that are actually same (because you only need to rotate cube on vertical axis to get same position again.... bcda is same as abcd rotated by 90 degrees). So the actual combinations in which 4 sides can be arranged = 4! / 4 = 6

So top and bottom can be arranged in 30 ways and or each such way the other 4 sides can be arranged in 6 ways. So total = 30*6 = 180

However, for any top/bottom combo of ab and sides of cdef, if you reverse the cube (top, bottom = ba), there is another sides combination that would look the same (bafedc). Hence the TOTAL = 180 / 2 = 90

What am I doing wrong? please help...[/list]
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by emalby » Mon May 16, 2011 7:51 am
taken from another website:
Let the six faces be a, b, c, d, e, and f. With face a opposite face b there are six arrangements for the other four colours around the cube: cdef, cdfe, cedf, cefd, cfde and cfed. Likewise for the face a opposite face c ; face a opposite face d ; face a opposite face e ; and face a opposite face f. All have six arrangements for the remaining five colours. Hence the total is 5 x 6 = 30 arrangements.
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by muthu8622 » Wed May 18, 2011 11:00 pm
I was thinking of the question for this few days and i came to this solution.. hope its accepted or please let me know if i'm wrong..

since there is 6 faces therefore there will be 6! ways to colour all the 6 faces.. its also 3D object and a cube has 8 rotation ( i'm using the concept of circular permutation in 3D) and in every vertex there will be 3 faces and there will be 3 cicular permutation

Hence, i put this way 6!/(8x3) = 720/24 = 30 ways

Is this accepted? or wrong?
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by tgou008 » Sun May 22, 2011 2:29 pm
abhi332 wrote:You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How
many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to
look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

[spoiler]OA:B[/spoiler]

I actually solved this using combinatorics.

If repetition was not an issue (i.e., the stmt in brackets above did not appear in the question) then the answer would simply be 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

However, as repetitions (aided by rotation) are not to count towards our total the colours are limited by the number of possible rotations for each face; namely, 4.
Therefore, the combinatoric is now 6!/4! = 6 x 5 = 30
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by giovanni.gastone » Wed May 25, 2011 1:10 pm
tgou008 wrote:
abhi332 wrote:You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How
many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to
look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

[spoiler]OA:B[/spoiler]

I actually solved this using combinatorics.

If repetition was not an issue (i.e., the stmt in brackets above did not appear in the question) then the answer would simply be 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

However, as repetitions (aided by rotation) are not to count towards our total the colours are limited by the number of possible rotations for each face; namely, 4.
Therefore, the combinatoric is now 6!/4! = 6 x 5 = 30
Hi tgou008, can you elaborate on this bit more?

"the colours are limited by the number of possible rotations for each face; namely, 4."

Why are there only 4 rotations for each color, and how does this translate into 4!?

Thank you!

Gio
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by chiru1849 » Sun Jun 12, 2011 9:03 am
The cube has six sides. Lets assign a number to each side from 1 through 6 with sides 5 and 6 being opposite to each other and sides 1 to 4 forming the four side faces. Let's fix side 1 as the referrent side. Having fixed side one sides 2, 3 and 4 can take 5, 4 and 3 colours respectively. Therefore 5*4*3 = 60. The other two sides take the remaining two colours but the cube will still look the same even if the colours on the teo faces are interchanged. Therefore the number of unique coloured cubes that you can get is 60 and therefore choice D.
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by Rastis » Fri Jan 06, 2012 11:14 am
Anyone able to provide an easy to understand explanation of how to do this problem?
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by santhoshsram » Fri Jan 06, 2012 5:23 pm
Found this on a different forum. Hope this helps.

https://www.techinterview.org/post/518750698/cube
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by ronnie1985 » Sun Apr 01, 2012 6:34 am
The answer is 6!/6 = 5! = 120
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