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    prob.

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    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
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    I think the answer choice is E (27 ways). Because the paint-color that is put on one cube is independent of the color put on the rest of the two cubes.

    Therefore, the total no. of ways of painting any of the given 3 colors on a cube is 3 ways. Since there are 3 cubes, the total ways = 3^3 = 3*3*3 = 27.

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    It should be 27

    Here is how
    If all cubes are painted same colour then they can be painter in 3 ways
    If each one painted in diff colours they can be painted in 6 ways
    If two are painted in one colour and the third one in diff colour they can be painted in 18 ways

    Hence total of 27 ways
    Elton

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    Quote Originally Posted by ittakes2win View Post
    I think the answer choice is E (27 ways). Because the paint-color that is put on one cube is independent of the color put on the rest of the two cubes.

    Therefore, the total no. of ways of painting any of the given 3 colors on a cube is 3 ways. Since there are 3 cubes, the total ways = 3^3 = 3*3*3 = 27.
    I have different thought here. The answer 27 is the result of considering order difference. But the question clarified oder is NOT important. My opinion is 10 ways.

    All 3 colors used - 1 way
    1 color -- 3 ways
    2 colors -- 2*(3C2 )= 6 ways

    Total is 10 ways

    What's the Official Answer?

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    Yes, D must be the answer. It is my fault that I ignored the additional info. provided in the problem statement completely. Thanks for correcting me.

    It is like this:

    Total no. of ways of painting all the 3 cubes with the same color (R,B or G) is: GGG, RRR, BBB - 3 ways

    Total no. of ways of painting all the cubes with a different color combinations is:
    RGB, RBG, BRG, BGR, GBR, GRB - 6 ways. However, with the order info. given, all of these 6 ways can be assumed to be just 1 way.

    Total no. of ways of painting 2 cubes with one color and the 3rd one with a different color is:
    GGB, GBG, BGG - 3ways
    GGR, GRG, RGG - 3ways
    BBG, BGB, GBB - 3ways
    BBR, BRB, RBB - 3ways
    RRG, RGR, GRR - 3ways
    RRB, RBR, BRR - 3 ways.
    Total ways - 3+3+3+3+3+3 = 18 ways.
    Again with the order info. given, all of these 18 ways can be assumed to be 6 unique ways.

    Therefore, total no. of ways = 1 + 3 + 6 = 10 ways.

    However, if the order of the color combinations were not to be ignored then it would be E (27 ways).

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    You don't need all those calculations.


    3^3=27 D

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    pls. post the Official Answer?

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    Official Answer is D (10)
    Thanks
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