Probability when painting cubes

[Roni1985]

Homework Statement

How many different ways can 3 cubes be pained if each cube is painted one color and only 3 colors red, blue, and green are available ? ( order is not considered, for example, green, green, blue is considered the same as green, blue, green).

Homework Equations

The Attempt at a Solution

I couldn't recall the way to get rid or repeating outcomes.
I tried to compute them one by one and I got 10, but how is it using faster ways ?


Thanks.

 

[CompuChip]

I would split the calculation in the number of distinct colors used.

[3 colors] If every cube has a different color, a quick count gives 3! = 6 possible colorings, but that doesn't take the order into account. How many possibilities are there actually?
[2 colors] If two cubes have the same color, you can choose 3 colors for one and 2 for another, which gives 6 possibilities, but there are how many different orderings (e.g. in how many ways can you order aab).
[1 color] If only one color is used, all cubes have this color, so there are 3 possibilities (i.e. all red, all green or all blue).

When I add this all up (or write them all out, just to check it) I get something even smaller than 10.

 

[Roni1985]

I would split the calculation in the number of distinct colors used.

[3 colors] If every cube has a different color, a quick count gives 3! = 6 possible colorings, but that doesn't take the order into account. How many possibilities are there actually?
[2 colors] If two cubes have the same color, you can choose 3 colors for one and 2 for another, which gives 6 possibilities, but there are how many different orderings (e.g. in how many ways can you order aab).
[1 color] If only one color is used, all cubes have this color, so there are 3 possibilities (i.e. all red, all green or all blue).

When I add this all up (or write them all out, just to check it) I get something even smaller than 10.

Hi,

How do you get something smaller than 10 ?
RRR BBB GGG
RRB BBR GGR
RRG BBG GGB
RGB

Exactly 10.

So your method is basically also listing them, right ?
I thought there was a shortcut to solve this one.

 

[╔(σ_σ)╝]

The way I would solve this is using combinations.

{3 \choose 0} + {3 \choose 1} +{3 \choose 2} = 10

Basically, you count the numbers of ways you can choose the combinations without order.

 

[gomunkul51]

@╔(σ_σ)╝

Please explain how you got that equation.

 

[╔(σ_σ)╝]

@╔(σ_σ)╝

Please explain how you got that equation.

I bascially counted the number of distinct ways to pick the colours on the balls given you have not pick any ball, i have pick one of the three colours and I have picked two of the three colours.

 

[gomunkul51]

yup, I don't remember much from my intro into prob. & stat. course :)

corecct me if I'm wrong:

<br /> {3 \choose 0} + {3 \choose 1} +{3 \choose 2} = 10<br />

<br /> 1 + 3 + 3 = 7<br />

not 10 ?
I solved it semi-hard way, by using the multiplication rule:

1.1.1 = 1 (choosing RBG without repetition)
3.1.1 = 3 (choosing 3 same colors)
3.1.2 = 6 (choosing 2 of the same color and 1 other)
= 10

 

[╔(σ_σ)╝]

yup, I don't remember much from my intro into prob. & stat. course :)

corecct me if I'm wrong:

<br /> {3 \choose 0} + {3 \choose 1} +{3 \choose 2} = 10<br />

<br /> 1 + 3 + 3 = 7<br />

not 10 ?



I solved it semi-hard way, by using the multiplication rule:

1.1.1 = 1 (choosing RBG without repetition)
3.1.1 = 3 (choosing 3 same colors)
3.1.2 = 6 (choosing 2 of the same color and 1 other)
= 10

I forgot how to add .


Clearly, that approach was incorrect.

Sry OP.

 

[CompuChip]

Yes, I fell for that too :-)