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combinations

This topic has 2 expert replies and 3 member replies
Nidhs Senior | Next Rank: 100 Posts Default Avatar
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combinations

Post Wed Jan 30, 2008 7:35 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    How many different ways can 3 cubes be painted if each cube is painted one color and only 3 colors red, blue, green are available? ( order is not considered, for example green, green blue is condidered the same as green, blue, green.)
    a)2 b)3 c) 9 d)10 e) 27

    can someone please expain to me a shorter method of solving this other than listing the orders down

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    Stuart Kovinsky GMAT Instructor
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    Post Wed Jan 30, 2008 7:52 pm
    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.

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    sibbineni Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Jan 30, 2008 7:56 pm
    I agree with stuart

    ayushiiitm Master | Next Rank: 500 Posts
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    Post Tue Jun 29, 2010 2:05 am
    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

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    Stuart Kovinsky GMAT Instructor
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    Post Tue Jun 29, 2010 9:21 am
    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.

    Thanked by: ayushiiitm
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    ayushiiitm Master | Next Rank: 500 Posts
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    Post Tue Jun 29, 2010 9:35 am
    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

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