Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

09 Sep 2013, 06:45

12

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

39% (02:01) correct
61% (01:30) wrong based on 152 sessions

HideShow timer Statistics

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]

Show Tags

10 Sep 2013, 06:16

3

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

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]

Show Tags

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]

Show Tags

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]

Show Tags

28 Dec 2013, 03:56

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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]

Show Tags

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]

Show Tags

28 May 2014, 03:41

2

This post received KUDOS

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]

Show Tags

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? _________________

The Mind is everything . What you think you become. - Lord Buddha

Consider giving KUDOS if you appreciate my post !!

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

Show Tags

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]

Show Tags

15 Jan 2015, 23:29

1

This post received KUDOS

Expert's post

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. _________________

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

Show Tags

12 May 2016, 23:40

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...