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Combinations

by Woozler » Mon Dec 27, 2010 8:01 pm
If we have 5 red, 4 white, 3 black and 2 yellow marbles, how many different 2 marble combinations are there? This is a really basic combinatorics question, yet I'm blanking. What is the general formula for that? The Manhattan GMAT book was not particularly thorough with this.

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by anu009 » Mon Dec 27, 2010 8:17 pm
Hello

The combination formula goes nCr where it is equal to n!/r!(n-r)!

If you subsitute the values of picking r itmes from a set of n that is 2 marbles from a set of 4 you will get 6 as the result.
Let me know what are the answer choices and if this is the correct solution.

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by Woozler » Mon Dec 27, 2010 8:22 pm
But why are we paying no attention to the fact that there are 4 different colors?
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by Anurag@Gurome » Mon Dec 27, 2010 8:43 pm
Woozler wrote:If we have 5 red, 4 white, 3 black and 2 yellow marbles, how many different 2 marble combinations are there?
The number of "different 2 marble combinations" is = 4C2 = 6

Why is so?
Because a red-red pair is same as another red-red pair. Also a red-black pair is same as another red-black pair. Only way we can get "different 2 marble combination" if we consider each of them of different color.

Therefore the problem reduces to: In how many ways 2 colors can be selected out of 4?
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