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

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.

Similar question to practice: a-cube-marked-1-2-3-4-5-and-6-on-its-six-faces-three-89198.html
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The number of arrangements of n distinct objects in a row is given by \(n!\).
The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).

From Gmat Club Math Book (combinatorics chapter):
"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:

\(R = \frac{n!}{n} = (n-1)!\)"

Questions about this concept to practice:
seven-men-and-seven-women-have-to-sit-around-a-circular-92402.html
a-group-of-four-women-and-three-men-have-tickets-for-seven-a-88604.html
the-number-of-ways-in-which-5-men-and-6-women-can-be-seated-94915.html
in-how-many-different-ways-can-4-ladies-and-4-gentlemen-be-102187.html
4-couples-are-seating-at-a-round-tables-how-many-ways-can-131048.html
at-a-party-5-people-are-to-be-seated-around-a-circular-104101.html
seven-family-members-are-seated-around-their-circular-dinner-102184.html
seven-men-and-seven-women-have-to-sit-around-a-circular-11473.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
seven-men-and-five-women-have-to-sit-around-a-circular-table-98185.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
find-the-number-of-ways-in-which-four-men-two-women-and-a-106919.html
gmat-club-monday-giveaway-155157.html (700+)



Hope this helps.
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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:

_
_ _ _ _
_

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).

Posted from my mobile device

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.

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

Fin27

Think of circular arrangement. When you place the first person, and make him sit on any seat, do you choose a person? No. You just pick anyone and put him anywhere. After that all seats become distinct. You could start with any person, it won't matter. What matters is how they are placed relative to each other, not in which order you made them sit. Whether A sat first or B sat first is irrelevant. At the end of it, it is only about who is sitting where relative to each other.

Similarly here, initially, all 6 faces are identical. You put any one colour on any face in 1 way. Now your opposite face is distinct but the other 4 faces are still identical.

Because after you select a face and put a colour on it, the face opposite to it becomes distinct.
But all the other 4 faces on the SIDES are still identical. Look at a cube to understand this. If you have colours on top and bottom faces and no colour or same colour on all side faces, all side faces are identical.

Which one you start with doesn't matter. So you again place a colour on any one of the 4 faces in 1 way. Now the other 3 faces are distinct - one is opposite to the coloured face, one is on the right and one is on the left.

In ciurlar arrangements, the moment you place the first person, all other seats become distinct.

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

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?

@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.

Sent from my Pixel XL using GMAT Club Forum mobile app

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No, this question is fine.
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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.
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While painting the top face..why don't we count the number of ways of choosing a colour.

Posted from my mobile device

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