jlgdr wrote:

Bunuel wrote:

chetan86 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

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.htmlI 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:

_

_ _ _ _

_

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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