Combinatrics

[rijul007]

In how many ways a cube can be colored using 6 colors if each color should be used atleast once?

A. 720
B. 120
C. 20
D. 30
E. 60

[bijoyajj]

In how many ways a cube can be colored using 6 colors if each color should be used atleast once?

A. 720
B. 120
C. 20
D. 30
E. 60

6! ways (6*5*4*3*2*1 - since color once used can't be used for other sides).. IMO a

[rijul007]

I am afraid thats not the right ans bijoyajj


The answer is Option D

Explainations plz...

[rijul007]

My approach:

6 sides of the cube are identical

so we can fix any color any on any side, doesnt matter
now opposite to that side of the cube any of the other 5 colors can be painted
leaving us with four colors and 4 walls [ similar to arranging 4 men on 4 chairs round a circular table] => (4-1)! = 6


So total no of arrangements = 6*5 = 30


Knowing the correct option beforehand forced me to think this way and arrive to the answer.
Is this the correct way to solve the ques???

[GmatMathPro]

You have to take into account rotational symmetries.

Imagine any given valid way to paint the cube with a different color on each of the six faces. Now, if you put the cube on a table, you could put any of the 6 faces facing up, and you could rotate it 90 degrees 4 times around a vertical axis that connects the top and bottom faces. That's 6*4=24 different arrangements for that one way of painting it. So, when you do 6*5*4*3*2*1=720, that is going to count each unique way of painting it 24 times. Hence, we have to divide by 24. 720/24=30.

Ans: D

[pemdas]

surprised with d thought it's e
with 6 colors applied individually it will make 6 ways
then 2 colors can be applied out of 6 colors, C(6,2) or 15 ways
additionally 3,4,5 and 6 out of 6 colors can be applied with the combination sets of 20,15,3 and 1 ways respectively.

in total 60 ways (6+15+20+15+3+1)

In how many ways a cube can be colored using 6 colors if each color should be used atleast once?

A. 720
B. 120
C. 20
D. 30
E. 60

[GmatMathPro]

surprised with d thought it's e
with 6 colors applied individually it will make 6 ways
then 2 colors can be applied out of 6 colors, C(6,2) or 15 ways
additionally 3,4,5 and 6 out of 6 colors can be applied with the combination sets of 20,15,3 and 1 ways respectively.

in total 60 ways (6+15+20+15+3+1)

If I understand your solution, it doesn't look like you're including the constraint that every color has to be used at least once. Also, there would be more than one way to arrange the colors in most cases once you choose them.

[smackmartine]

IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36
However, out of these 36 ways we have to exclude the cases when the cube is colored with a single color..i.e in 6 ways

So , required # of ways = 36-6 = 30

[GmatMathPro]

IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36

Why 6*6?

[pemdas]

Imagine any given valid way to paint the cube with a different color on each of the six faces.

different interpretations are possible with different colors on the cube's sides

So, when you do 6*5*4*3*2*1=720, that is going to count each unique way of painting it 24 times.

yes, we are counting unique ways as we perform simple permutation here of each color among the six given P(6,1). I am not sure if this is useful here as the colors put in different orders was not accepted by me. I voted for the sets of different colors (1,2,3,4,5,6) selected out of possible 6 colors. But even then with permuting each color out of 6 we get 720 and answer A.

Now, if you put the cube on a table, you could put any of the 6 faces facing up, and you could rotate it 90 degrees 4 times around a vertical axis that connects the top and bottom faces. That's 6*4=24 different arrangements for that one way of painting it. Hence, we have to divide by 24. 720/24=30.

Cannot agree with this more. As it was correctly noted before, there are 720 unique arrangements and when we discount these arrangements by 24 we are losing some valid sets here.

My take on this question with the constraint "each color is used at least once" is that does not mean each color must be used each time other colors are applied. I use each color one and more than one times and find the sets of different color combination.

[smackmartine]

IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36

Why 6*6?

Pete, I applied the same concept: if a work that can be done in m different ways and another work can be done is n different ways, total no. of ways both of them can be done in m*n ways.
eg. to prepare a meal there are following items - 3 kinds of breads , 3 kinds of toppings and 2 kinds of drinks..so there can be 3.3.2 kinds of meal. Is n't this right?

[GmatMathPro]

Okay, I just want to make sure I'm clear on what you're saying. Please let me know if I misinterpreted anything. So you're saying that the problem is essentially choosing which of the 6 colors are used to paint the cube, that the different orderings of the colors does not matter, and that you can repeat the same color on different faces of the cube.

So, if the six colors are red, orange, yellow, green, purple, and blue, with each color used at least once....

You would consider 2 blue faces, 2 orange faces, and 2 yellow faces as a valid way to color the cube?

You would consider 5 blue faces and 1 orange face as the same as 3 blue faces and 3 orange faces?

You would consider all different arrangements of 6 distinct colors to be the same, so that, for example, having orange across from blue would be the same as having orange adjacent to blue?

Finally, to be clear, I did not say that there are 720 unique ways to paint the cube, but that there are 30 unique ways to paint the cube, and 6*5*4*3*2*1 counts each of those ways 24 times.

[GmatMathPro]

IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36

Why 6*6?

Pete, I applied the same concept: if a work that can be done in m different ways and another work can be done is n different ways, total no. of ways both of them can be done in m*n ways.
eg. to prepare a meal there are following items - 3 kinds of breads , 3 kinds of toppings and 2 kinds of drinks..so there can be 3.3.2 kinds of meal. Is n't this right?

But exactly what two things are you saying can be done in 6 ways each? 6 ways to choose a side of the cube and then 6 ways to choose a color? If so, that would only get one side painted, and only 6 of these 36 paintings would be unique, due to rotational symmetry.

Also, if your plan is to not worry about the constraint that each color should be used at least once until the end, it would make more sense to say, "there are 6 ways to choose the color for the first side, times 6 ways to choose the color for the second side, times 6 ways to choose the color for the third side,...." and so on, so there would be 6^6=46656 ways to paint the cube. But that would include many more invalid colorings than just the ones where all faces are the same color. It would also include the ones that use one color 5 times, 4 times, 3 times, and 2 times, plus tons of repeats from rotational symmetry. So you would have a lot more stuff to subtract than just 6. It would get pretty complicated counting up what to subtract out, so it's better to start with the constraint in place and say that there are 6*5*4*3*2*1 ways to choose a color for each side and then divide out the repeats.

Or, you could set it up the way rijul007 did, and you wouldn't have to worry so much about all the possible rotations.

[rijul007]

okay, so my ans was right too

thanx Pete for making it much more clearer.

[shankar.ashwin]

I posted this question few days back and Mitch had a good solution too,
https://www.beatthegmat.com/simple-and-t ... 93158.html