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There are 3 red chips and 2 blue ones. When arranged in a

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There are 3 red chips and 2 blue ones. When arranged in a [#permalink] New post 25 Oct 2007, 09:20
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Difficulty:

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Question Stats:

64% (01:35) correct 36% (01:09) wrong based on 283 sessions
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: m09-72702.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 15:14, edited 2 times in total.
Added the OA.
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Re: Combinations [#permalink] New post 27 Oct 2007, 15:12
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 [#permalink] New post 27 Nov 2007, 00:22
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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 [#permalink] New post 27 Nov 2007, 01:55
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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 [#permalink] New post 22 Dec 2007, 02:26
Hi guys,
in this question the order is important. But will not RRRB1B2 be the same as RRRB2B1?
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 [#permalink] New post 22 Dec 2007, 06:13
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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Re: [#permalink] New post 27 Jan 2008, 12:49
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!*(5-3)! but rather a straight factorial of the two different sets of chips? what is the rule on this?
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Re: Combinations [#permalink] New post 27 Jan 2008, 12:58
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 [#permalink] New post 18 Mar 2008, 11:34
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 [#permalink] New post 18 Mar 2008, 21:04
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 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

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Re: Combinations [#permalink] New post 19 Mar 2008, 21:07
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 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?
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Re: Combinations [#permalink] New post 19 Mar 2008, 21:17
Expert's post
Thu Oct 25, 2007 7:27 pm versus
Fri Oct 26, 2007 7:26 am

tricky GMAT TIGER :)
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Re: Combinations [#permalink] New post 19 Mar 2008, 22:41
walker wrote:
Thu Oct 25, 2007 7:27 pm versus
Fri Oct 26, 2007 7:26 am

tricky GMAT TIGER :)


foolish
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me.
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Re: Combinations [#permalink] New post 25 Jul 2009, 10:02
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its 5C3 x 2C2 = 10 x 1 = 10
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Re: Combinations [#permalink] New post 24 Sep 2009, 19:50
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 [#permalink] New post 26 Sep 2009, 12:41
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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 [#permalink] New post 27 Sep 2009, 01:37
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 [#permalink] New post 24 Dec 2009, 09:23
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 [#permalink] New post 24 Dec 2009, 09:44
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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 [#permalink] New post 31 Dec 2009, 09:12
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
Re: Combinations   [#permalink] 31 Dec 2009, 09:12
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