0
$\begingroup$

In how many ways can you color a cube if you have to color four sides with four different colors (i.e. green, blue, black and red) and other two sides with same color (i.e. yellow).

Assume the sides are indistinguishable.

$\endgroup$

closed as off-topic by Rory Daulton, user147263, Mario Carneiro, Servaes, yoknapatawpha Oct 24 '15 at 22:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Community, Mario Carneiro, Servaes, yoknapatawpha
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hint: In how many ways can you choose which $2$ of the $6$ sides are yellow? In how many ways can you choose to color the four remaining sides? $\endgroup$ – Marcus M Oct 24 '15 at 17:35
  • $\begingroup$ what is your current understanding of combinatorics? can you outline your attempts in more detail? $\endgroup$ – user264985 Oct 24 '15 at 17:41
2
$\begingroup$

Label the faces, then you have $\frac{6!}{2}$ ways of colouring the faces. But you can orientate the cube in $6\times 4$ ways, so the answer is $$\frac{5!}{8}$$

$\endgroup$
  • $\begingroup$ Can you explain how do we have 6!/2 ways of coloring faces? $\endgroup$ – Infinity Oct 24 '15 at 17:50
  • 1
    $\begingroup$ If you label the faces then its the same as counting the ways that the colours can be laid out in a line. So we have 6 choices for the first, 5 for the second... =6!. But as there are 2 identical (the yellow ones) we have counted each permutation twice, so divide by 2. $\endgroup$ – Chris Kerridge Oct 24 '15 at 17:53
  • $\begingroup$ You can't label the faces. The six faces are identical. $\endgroup$ – Infinity Oct 24 '15 at 17:53
  • $\begingroup$ Assume the sides are indistinguishable. $\endgroup$ – Infinity Oct 24 '15 at 17:54
  • $\begingroup$ He's labeling the faces as a tool to count the colorings. Then he divides by the number of possible ways of labeling the faces. $\endgroup$ – Dustan Levenstein Oct 24 '15 at 17:54
2
$\begingroup$

Assuming the orientation of the cube is fixed, you can count the colorings by first choosing the two yellow sides, then permuting the remaining four sides.

$$\binom{6}{2} \cdot 4! = 15 \cdot 24 = 360.$$

If, as you stated in the edit, you are allowing the cube to be rotated any way, then you divide by the number of ways of orientating the cube, as Chris Kerridge has done. This approach requires the use of the fact that there is no way to reorient a colored cube and have it look the same as before.

$\endgroup$
  • 1
    $\begingroup$ There is a flaw in this explanation. I can rotate a cube in 3d space. Suppose I choose two opposite sides as yellow. Now according to you: there are 4! ways to color remaining 4 sides but actually there are only 3! as the sides can occur in cyclic order like (green,black,blue,red) or (black,blue,red,green) or (blue red green black)... All of these give us only one way to color the cube. $\endgroup$ – Infinity Oct 24 '15 at 17:39
  • 1
    $\begingroup$ So you're asking to count them assuming the sides are indistinguishable. It might've been a good idea to include that in the question. $\endgroup$ – Dustan Levenstein Oct 24 '15 at 17:41
2
$\begingroup$

There are two possibilities: (i) the yellows are on opposite faces or (ii) they are not.

(i) Put the cube on a table, a yellow side down. Rotate until black is facing you. There are $3$ ways to choose the colour of the face opposite the black. Now we have no more options, for one of the remaining colourings can be turned into the other by putting the other yellow face down.

(ii) There are $\binom{4}{2}=6$ ways to choose the set of two colours facing the yellow sides. Put the cube on the table, with a yellow face down and, of the two colours chosen for the opposite side, the one earlier in the alphabet on top. Rotate until the other yellow is facing you. There are $2$ choices for the remaining colours, for a total of $12$.

Add up. We get $15$.

$\endgroup$
  • $\begingroup$ I like your approach, but how is that you don't get more than $15$ possibilities. According to Dustun Levenstein and Chris Kerridge, we get $45$. $$\frac{4!}{2^3} \binom{6}{2}$$ $\endgroup$ – hlapointe Oct 25 '15 at 15:31
  • $\begingroup$ I think you're commit a mistake, because in (i) you have more than $3$ ways to arrange the color. If I'm wrong, can you explain it to me. $\endgroup$ – hlapointe Oct 25 '15 at 16:00
  • 1
    $\begingroup$ Actually, Chris Kerridge also has a total count of $15$. The question is of a srandard kind, counting "colourings" modulo a group of symmetries, here the rigid motions. And it turns out there are very few truly distinct colourings of what I called Type (i), I wrote in effect that (up to rigid motions) there is only one colouring with the yellows opposite and green opposite black. To show that is not true, one has to exhibit two. $\endgroup$ – André Nicolas Oct 25 '15 at 16:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.