How many different ways can 3 cubes be painted if each cube is painted one color and only the 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.)

(A) 2
(B) 3
(C) 9
(D) 10
(E) 27

I think it should be E 27
3 diff ways to paint the first cube * 3 diff ways for 2nd cube *3 for the last

but below is a different explanation
For all colours, there is one case when all cubes will be painted with unique colours i.e. R,B,G.

Now, for any colour, lets say B; possible ways of painting the cubes are BBB, BB(R/G). Thus for one colour there are 1+2=3 ways.

Take one cube (the "first cube"), you can colour this in 3 different ways.

Suppose we colour this first cube red, then there are 3 different choices for the second cube, and 3 for the last cube. This gives a total of 9 choices, with the first cube red. Now some of these coincide (by "permuting" the cubes - which means "moving them around"), for example R,B,G and R,G,B are the same. The choices are (consider a "tree like diagram" - if you want me to ellaborate on this i will) RRR, RBR, RBB, RGG,RGR and RGB... 6 in total.

Now if we choose our first cube as blue, then our second cube can take on 3 different colours. But the colour of this second cube cannot be red, because then the final cubes colour will (by permuting the cubes) give a colour that was already done in the above case (where we choose red as the colour of the first cube). Thus we have two choices for the colour of the second cube: green or blue. Now the colour of the last cube can also take on two colours only (namely blue or green), since chosing the colour red (by permuting the cubes) will give a choice of colours already given in the stage where we chose red as the colour of the first cube. Thus, for blue as the colour of the first cube, we have 2*2=4 different choices (Green and Blue, Blue and Green, Blue and Blue and Green and Green). This gives us a total of 3 choices for the 3 cubes colours (with blue as the colour of the first cube): BGB BGG, BBB.

Finally lets let the first cube have colour green. We cannot choose the second cube to have colour red (because, by permuting, this combination will, on chosing a colour for the last cube, be a colour covered in the first paragraph), and similarly we can't let this second cube have colour blue (because this will, by permuting, lead to a case covered in the second paragraph). Thus the second cube must have colour green. Similarly the third cube must have colour green. Thus, with green as the colour of the first cube, we can only have 1 choice for the colours of the other cubes - Green and Green. This gives the option GGG.

Therefore there are: 6 + 3 + 1 = 10 choices for colourings.

[I'm going to read over this and check that I haven't made a mistake with cases, but perhaps people want to read this and give me some feedback]

Hope this helps! If not I hope it points you in the right direction...

Secondly, to explain why your answer of 27 was wrong... here we go...

Your saying when you choose your first cube, you then have 3 choices for the second cubes colour, and then when this is chosen you have a further 3 choices for the 3rd cubes colour, giving a total of 3*3*3=27 choices.

However, if we choose the first cubes colour as R, we are then free to choose B as the second and G as the third. But likewise, under your assumption, we are allowed to choose the second as G and the third as B. This means you have accounted for choices RBG and RGB, which the question you quoted says should be regarded as the same.

27 would be the answer if we had 3 cubes, 3 colours, and ordering DID matter. It would also make this whole thing a lot easier!