Permutations and combinations (1 Viewer)

[fire and ice]

Hey guys,

Does anyone know how to do this question

'The six faces of a number of identical cubes are painted in six distinct colors. How many different cubes can be formed?'

The answer is 120 but i don't know how to get 120.

Any help is appreciated.

 

[braintic]

Hey guys,

Does anyone know how to do this question

'The six faces of a number of identical cubes are painted in six distinct colors. How many different cubes can be formed?'

The answer is 120 but i don't know how to get 120.

Any help is appreciated.

I'm getting 30. Was this from a book, or made up by your teacher?

 

[superSAIyan2]

so the dice has the numbers 1,2,3,4,5 and 6
now we have to colour each number in a different colour (from 6 available colours)
so you can paint the first number in 6 ways
you can paint the second numbber in 5 ways
can choose 4 colours to paint the 3rd number

from this the total number of possibilities is 6x5x4x3x2x1 = 6! =120

another explanation is that since this is an ordered selection : you are arranging 6 different numbers to 6 different colours the total number of arrangements possible is 6P6 = 6! = 120

 

[fire and ice]

It was from the textbook cambridge yr 12

 

[fire and ice]

wait the answer was 30, my mistake. I'm very sorry superSAIyan2

 

[braintic]

so the dice has the numbers 1,2,3,4,5 and 6
now we have to colour each number in a different colour (from 6 available colours)
so you can paint the first number in 6 ways
you can paint the second numbber in 5 ways
can choose 4 colours to paint the 3rd number

from this the total number of possibilities is 6x5x4x3x2x1 = 6! =120

another explanation is that since this is an ordered selection : you are arranging 6 different numbers to 6 different colours the total number of arrangements possible is 6P6 = 6! = 120

But 6! = 720

 

[braintic]

wait the answer was 30, my mistake. I'm very sorry superSAIyan2

Do you still need an explanation?

 

[Sy123]

so the dice has the numbers 1,2,3,4,5 and 6
now we have to colour each number in a different colour (from 6 available colours)
so you can paint the first number in 6 ways
you can paint the second numbber in 5 ways
can choose 4 colours to paint the 3rd number

from this the total number of possibilities is 6x5x4x3x2x1 = 6! =120

another explanation is that since this is an ordered selection : you are arranging 6 different numbers to 6 different colours the total number of arrangements possible is 6P6 = 6! = 120

Its asking for unique cubes, so lets say we have our cube where red is top and white is bottom.
Another arrangement (according to your solution) is white on top red on bottom, but they are the same cube.

First lets arrange the top and bottom faces of the cube.

We pick 2 colours from the 6, so that is ways to do so.

we still need to colour in the sides, there are 4 colours remaining. Now for these 4 colours, a, b, c, d.
It is similar to arranging around a round table, the total arrangements is (4-1)! = 3! = 6
For example, looking at the cube from the top (the net)
The four sides could be:

...a
b.....c
...d

...a
c.....b
...d

Can you see how these 2 arrangements are the same if we put it in a cube?

I have no elegant way of doing this, but first draw out the 6 cases that we can have for the sides. Then eliminate all that is not unique.
You will have 2 remaining.

From the beginning, 6C2 * 2 = 30

 

[fire and ice]

Thank you, it makes so much sense now

 

[braintic]

An easier way to think about it:
Place the first colour on any face
There are five ways to choose its opposite colour.
What is left amounts to arranging 4 objects in a circle: 3! possibilities.
Total possibilities = 5 times 3! = 30

 

[superSAIyan2]

oh shoot did 5! by mistake. Lol and i forgot to account for the fact that the dice can be rotated to give the same pattern