You have a six-sided cube and six cans of paint, each a diff

Thanks Bunues, great explanation!!

Can you give more detail on how you determined what method to use to solve this problem? I have never heard of the "circular arrangement" technique you used.

This helps a great deal! Thank you Bunuel!

I really didn't get how you used circular arrangements here.

We have six different colors and 6 different sides.

How many ways can we paint the cube with each side a different color?

Shouldn't it be 6! ?

Please clarify
Great problem btw

Cheers
J

Painting a cube with six different colors is more complicated than arranging six distinct things in a row.

If you have six distinct colors, say ROYGBP, you could place them in the slots below in 6! ways. Why? Any of the 6 colors could go in the first slot, any of the five remaining could go in the second, and so on. So the total is 6*5*4*3*2*1=6!
_ _ _ _ _ _

Now the cube is a combination of slots and a circular arrangement. Here is my sophisticated diagram for that:

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_ _ _ _
_

The four slots between the top and bottom 'faces' actually wrap around the whole cube. So, let's say for the four 'slots' you choose ROYG.

ROYG =GROY = YGRO = OYGR = ROYG

All of these arrangements are the same because essentially each one is just a rotation of the cube by 90 degrees, not a different paint job.

So we treat the four 'slots' in the middle the same way we would a circular table. Hence the above solutions: you can choose any color for the top face, you can choose one of the five remaining colors for the bottom face (5 ways), and since the four middle faces are 'in a circle' they can be arranged (4-1)!=3! ways. So the total is 5*3!

The tricky part is that we don't count the ways in which we can choose the color of the first face, since every color is going to be chosen anyway. Essentially, you are finding the ways you can paint the other sides relative to one of painted sides. Otherwise you are including in your total the number of different ways you can look at the cube (which don't constitute a new paint job).

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Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).

Total = 5*6 = 30.

Answer: B.

@Bunuel

, why haven't the ways to paint the first (top)face have been counted, i.e 6C1 ways to choose from amongst the 6 diffferent paints?? And what does the wording provided in the question in the parentheses entails?

Hi Bunuel,
Need your help on this. I could understand the circular and bottom face things. that's 5*(4-1)!. But I am not able to understand that top part. If initially no sides are painted, then we could chose 1 among 6 paints right? so shouldn't we multiply 5*(4-1)! with 6, as we have six choices initially?

Why don't we consider the 6 ways in which we can color the top face?

Which face is the top face? All faces are identical. You pick any color and put it on any one side. This can be done in one way only. This is like placing the first person at a round table. All places are identical so place the first one can be put anywhere. Similarly, the first paint can be put on any face of the cube. Now you have a top face (which we have just painted) and a bottom face and 4 identical sides.

@Bunuel

What about this one? Is this a GMAT question? or this too is out of scope? I found a lot of difficulty solving this one..and got it wrong the first time.

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No, this question is fine.

Okay. Please suggest somewhere to read about these types of problems..and can you also explain any other method to approach this question? Combinatorics is kind of my strength, I don't want to miss questions such as these on the GMAT. Please help me.

This question does not represent any particular/special type of combinatorics questions - we use circular permutation there and that's it. Links to circular permutation questions are given above. Don't know what else to suggest. Hopefully more practice should help.

Responding to a pm:


Note that 6C2 is the number of ways in which you select 2 faces out of 6 DISTINCT faces. The case here is that you have 6 IDENTICAL faces. You pick any one for a colour of your choice in 1 way and call it TOP. Now you automatically have a fixed BOTTOM face. So there is no choosing. You choose a colour for it in 5 ways.
Next, you have 4 IDENTICAL sides. You pick any 1 in 1 way for a colour of your choice. The rest of the 3 faces are DISTINCT now so you can distribute the 3 colours to them in 3! ways.

So total ways = 5*3! = 30

While painting the top face..why don't we count the number of ways of choosing a colour.

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I have addressed it here: https://gmatclub.com/forum/you-have-a-s ... l#p1469980

Hi,

I am new in the forum and came across this question while practicing Permutations and Combinations. I went through the thread and understood the reasoning provided by everybody else.

However, I solved it using the following understanding. Bunuel please let me know if my understanding is justified:

We have 6 faces, so the top face can be any colour out of the 6 available colours. So we have 6 ways of chosing the colours.
Now, for the 4 faces we have 5 colours to chose from which we can do in 5C4 ways. This will give 5 different colour combinations for the 4 faces. The bottom face then would be whatever is left. So total no. of ways is 6*5c4 = 30

Thanks in Advance

This logic is not correct. Firstly, there is no top face for which you have 6 options. If you re-orient the cube, any face could be top face. Say if red is on top on my cube and it is on the side on yours, it is not necessary that our cubes are distinct. Perhaps you are just holding it that way. How will we know whether two cubes are distinct or not? We put red on top for both cubes. Now see whether we have the same colour at the bottom and at the sides in the same order.
Also, you have just selected 4 colours for the side out of 5. Their different placement will create different cubes. Say we have Yellow, blue, green and white for the sides. If yellow and blue are adjacent, this is different from the case in which yellow and blue are opposite to each other.