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But I was wondering that yesterday. when do you use xCx?
Can someone mention examples when and when not, as to be able to recognize the pattern? Thanks!

did any of you noticed that my post was posted at 9:26 pm on Thu Oct 25, 2007 and elgo's post also on the same day one minuet after mine. but his post is ahead of mine. I am wondering, why, though its not gmat off-topic? hmm........

GMAT TIGER wrote:

bmwhype2 wrote:

There are 3 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns?

did any of you noticed that my post was posted at 9:26 pm on Thu Oct 25, 2007 and elgo's post also on the same day one minuet after mine. but his post is ahead of mine. I am wondering, why, though its not gmat off-topic? hmm........

GMAT TIGER wrote:

bmwhype2 wrote:

There are 3 red chips and 2 blue ones. When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns?

10 12 24 60 100

= 5!/3!2! = 10

Is it too cold outside that people do not response, participate or discuss on a given topic?
_________________

I don't think we should use C here as basics say that Combination(C) means selection and P means arrangement.

Here we're looking to arrange 5 chips, so we should be using Permutation.

Total ways of arranging 5 chips= 5!.

Since 3 chips are of same color and 2 of same and there is no way to distinguish between those of same color, i.e., u can't say which one is b1 and whoch one is b2 as all of them are identical.

would be used if we were picking various combinations of picking (3 or 2 items out of a set of 5). Obviously I am wrong.

So that I can further grasp this concept - let me ask a twist on this question. If the set of 9 balls had 4 red, 3 green abd 2 blue - how do we determine all possible ways the 9 can be laid out?

The way this is solved will help me understand how the above simpler problem was solved.

would be used if we were picking various combinations of picking (3 or 2 items out of a set of 5). Obviously I am wrong.

So that I can further grasp this concept - let me ask a twist on this question. If the set of 9 balls had 4 red, 3 green abd 2 blue - how do we determine all possible ways the 9 can be laid out?

The way this is solved will help me understand how the above simpler problem was solved.

Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\(\frac{n!}{P1!*P2!*P3!*...*Pr!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is 6!/(2!2!), as there are 6 letters out of which g and o are represented twice.

In your example 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/(4!3!2!).

I believe the person was trying to understand how you get to an answer of 10.

Example is: There are 3 red chips and 2 blue ones, when arranged in a row, they form a certain color pattern, for example RBRRB. How many color patter. We know there are 5 chips = 5! there and 3! reds 2! blue. Therefore, 5!= (5*4*3*2*1)/(3! = 3*2*1) and (2!= 2*1) thus 5*4*3*2*1 / (3*2*1) (2*1) =5.4/2.1 =10

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