Combinatrics

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by pemdas » Tue Nov 01, 2011 12:51 am
actually after reading posts in this thread and going through the link-post provided by shankar I came to a different solution; if possible, please evaluate
the circular permutation of 6 colors applied onto the cube sides which can be rotated 4 times around its upper-bottom sides formed axis ==> (6-1)!

divided by 4 sides, likewise in a popular problem with bracelet applied with diamonds and flipped over ...

hence 5!/4=30

@Pete, am i missing something or it's equally good as yours?
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by GMATGuruNY » Tue Nov 01, 2011 1:57 am
Click on the link below for a solution that does not require a knowledge of rotational symmetry. The reasoning behind the solution is that for each pair of colors, there are only two options: the pair must be placed on two adjacent faces of the cube or on two opposite faces:

https://www.beatthegmat.com/simple-and-t ... 93158.html
Last edited by GMATGuruNY on Wed Nov 02, 2011 12:29 pm, edited 1 time in total.
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by GmatMathPro » Tue Nov 01, 2011 9:57 am
pemdas wrote: the circular permutation of 6 colors applied onto the cube sides which can be rotated 4 times around its upper-bottom sides formed axis ==> (6-1)!
I guess I don't see where (6-1)! comes from. In this part, are you trying to calculate the number of ways just the four vertical sides can be painted from six colors? If so, it should be (6*5*4*3)/4=90. (6-1)! would be like if you have 6 people sitting around a circular table with 6 seats. This is more like 6 people sitting around a circular table with 4 chairs. Or maybe I misunderstood you.
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