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

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You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 09 Sep 2013, 05:45
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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.)

(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
[Reveal] Spoiler: OA
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 09 Sep 2013, 07:06
Can someone explain why the answer is B?
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 10 Sep 2013, 03:43
Bunuel and other experts,

Can you pls suggest the best approach to solve this problem?
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 10 Sep 2013, 05:16
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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).

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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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 10 Sep 2013, 06:56
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).

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


Thanks Bunues, great explanation!!
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 27 Dec 2013, 19:12
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.

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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.
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 28 Dec 2013, 02:56
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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:

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+)

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html


Hope this helps.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 28 Dec 2013, 20:30
This helps a great deal! Thank you Bunuel!
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 27 May 2014, 05:19
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).

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


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
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 28 May 2014, 02:41
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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).

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


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).

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You have a six-sided cube and six cans of paint, each a diff [#permalink] New post 15 Oct 2014, 21:32
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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You have a six-sided cube and six cans of paint, each a diff   [#permalink] 15 Oct 2014, 21:32
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