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There are 3 red chips and 2 blue ones. When arranged in a
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Updated on: 09 Jul 2013, 16:14
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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? A. 10 B. 12 C. 24 D. 60 E. 100 OPEN DISCUSSION OF THIS QUESTIONS IS HERE: m0972702.html
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Originally posted by bmwhype2 on 25 Oct 2007, 10:20.
Last edited by Bunuel on 09 Jul 2013, 16:14, edited 2 times in total.
Added the OA.



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Re: Combinations
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27 Oct 2007, 16:12
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
wish i get something this easy on the real exam
5!/3!2!= 10



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Re: Combinations
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27 Nov 2007, 01:22
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
OA is A.
The OE uses 5C2.



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I got 10 as well. 5!/3!2!
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!



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Hi guys,
in this question the order is important. But will not RRRB1B2 be the same as RRRB2B1?



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CaspAreaGuy wrote: Hi guys, in this question the order is important. But will not RRRB1B2 be the same as RRRB2B1?
yup, so you use permutations with the number of multiples in the denominator.
5! = number of chips
3!2! = number of repeats (3 reds and 2 blues)
5!/3!2! = 10



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eschn3am wrote: CaspAreaGuy wrote: Hi guys, in this question the order is important. But will not RRRB1B2 be the same as RRRB2B1? yup, so you use permutations with the number of multiples in the denominator. 5! = number of chips 3!2! = number of repeats (3 reds and 2 blues) 5!/3!2! = 10 so the denominator in your fraction is not 3!*(53)! but rather a straight factorial of the two different sets of chips? what is the rule on this?



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Re: Combinations
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27 Jan 2008, 13:58
Using anagram method:
5_4_3_2_1 R_R_R_B_B
so.. 5!/Number of repeated letters (3!)(2!) = 10



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Re: Combinations
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18 Mar 2008, 12:34
jimmyjamesdonkey wrote: Using anagram method:
5_4_3_2_1 R_R_R_B_B
so.. 5!/Number of repeated letters (3!)(2!) = 10 the key is repetition: the denominator must show number of repeated elements, therefore 3!2!



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Re: Combinations
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18 Mar 2008, 22:04
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 offtopic? 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
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Re: Combinations
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19 Mar 2008, 22:07
GMAT TIGER wrote: 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 offtopic? 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?
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Re: Combinations
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19 Mar 2008, 22:17
Thu Oct 25, 2007 7:27 pm versus Fri Oct 26, 2007 7:26 am tricky GMAT TIGER
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Re: Combinations
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19 Mar 2008, 23:41
walker wrote: Thu Oct 25, 2007 7:27 pm versus Fri Oct 26, 2007 7:26 am tricky GMAT TIGER foolish . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . me.
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Re: Combinations
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25 Jul 2009, 11:02
its 5C3 x 2C2 = 10 x 1 = 10
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Re: Combinations
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24 Sep 2009, 20:50
Please correct me if I'm wrong at any point.
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.
So, the answer will be 5!/(3!*2!)



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Re: Combinations
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26 Sep 2009, 13:41
consider placing 3 things in 5 options...the rest 2 blues will get settled automatically... so 5C3 = 10 ways
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Re: Combinations
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27 Sep 2009, 02:37
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
Soln: Total number of patterns is 5!
Since 3 red chips are identical and 2 blue ones are identical thus we have = 5!/(2! * 3!) = 10 such different patterns



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Re: Combinations
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24 Dec 2009, 10:23
Pardon my ignorance.
I thought 5!/3!*2!
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.



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Re: Combinations
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24 Dec 2009, 10:44
junker wrote: Pardon my ignorance.
I thought 5!/3!*2!
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!). Hope it's clear.
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Re: Combinations
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31 Dec 2009, 10:12
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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