Consider 3 scenarios:

(1) all 3 the same colour... 3 options
(2) all 3 different colours... 1 option
(3) 2 of 1 colour and 1 of another colour (so we're using 2 of the 3 colours): 3C2 gives us the possible 2 colour choices. However, we need to double it, since if we choose R and B we could have RRB or RBB.

So, 3C2 * 2 = 3 * 2 = 6

3 + 1 + 6 = 10 possibilities.

I agree with stuart

Stuart Kovinsky wrote:Consider 3 scenarios:

(1) all 3 the same colour... 3 options
(2) all 3 different colours... 1 option
(3) 2 of 1 colour and 1 of another colour (so we're using 2 of the 3 colours): 3C2 gives us the possible 2 colour choices. However, we need to double it, since if we choose R and B we could have RRB or RBB.

So, 3C2 * 2 = 3 * 2 = 6

3 + 1 + 6 = 10 possibilities.

Should it not be 3*3*3

because we may say
for first cube we have 3 choice, for second we have 3 and third we have 3

ayushiiitm wrote:Should it not be 3*3*3

because we may say
for first cube we have 3 choice, for second we have 3 and third we have 3

Hi!

The problem with your approach is that you've included duplicates.

For example, you've counted:

RGB
RBG
GRB
GBR
BRG
BGR

as 6 different combinations, even thought they're all the same (one of each colour).

Since order doesn't matter in this question (we just want to know the colours of the cubes, we don't care in what order we painted them), you have to use a different counting method.

Stuart Kovinsky wrote:

ayushiiitm wrote:Should it not be 3*3*3

because we may say
for first cube we have 3 choice, for second we have 3 and third we have 3

Hi!

The problem with your approach is that you've included duplicates.

For example, you've counted:

RGB
RBG
GRB
GBR
BRG
BGR

as 6 different combinations, even thought they're all the same (one of each colour).

Since order doesn't matter in this question (we just want to know the colours of the cubes, we don't care in what order we painted them), you have to use a different counting method.

Thanks Stuart