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You have a six-sided cube and six cans of paint, each a diff [#permalink]

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09 Sep 2013, 06:45

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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.)

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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10 Sep 2013, 06:16

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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).

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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10 Sep 2013, 07:56

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).

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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27 Dec 2013, 20:12

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:

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.

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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28 Dec 2013, 03:56

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sjmarinov 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:

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.

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:

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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27 May 2014, 06:19

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).

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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28 May 2014, 03:41

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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).

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:

_ _ _ _ _ _

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).

You have a six-sided cube and six cans of paint, each a diff [#permalink]

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15 Oct 2014, 22:32

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.

@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? _________________

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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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15 Jan 2015, 22:57

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).

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?

Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]

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15 Jan 2015, 23:29

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Vinitkhicha1111 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.

@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?

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. _________________

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