combinations

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combinations

by Nidhs » Wed Jan 30, 2008 7:35 pm
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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by Stuart@KaplanGMAT » 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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by sibbineni » Wed Jan 30, 2008 7:56 pm
I agree with stuart

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by ayushiiitm » 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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by Stuart@KaplanGMAT » 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.
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by ayushiiitm » 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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