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

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CEO
Joined: 21 Jan 2007
Posts: 2581
Location: New York City
There are 3 red chips and 2 blue ones. When arranged in a  [#permalink]

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Updated on: 09 Jul 2013, 16:14
1
11
00:00

Difficulty:

25% (medium)

Question Stats:

65% (00:57) correct 35% (01:27) wrong based on 387 sessions

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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: m09-72702.html

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.
Senior Manager
Joined: 04 Jun 2007
Posts: 337

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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
CEO
Joined: 21 Jan 2007
Posts: 2581
Location: New York City

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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.
Senior Manager
Joined: 09 Oct 2007
Posts: 438

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27 Nov 2007, 02:55
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!
Manager
Joined: 01 Sep 2007
Posts: 92
Location: Astana

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22 Dec 2007, 03:26
Hi guys,
in this question the order is important. But will not RRRB1B2 be the same as RRRB2B1?
Director
Joined: 12 Jul 2007
Posts: 837

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22 Dec 2007, 07:13
2
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
Intern
Joined: 13 Jan 2008
Posts: 24

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27 Jan 2008, 13:49
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?
Director
Joined: 01 May 2007
Posts: 769

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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
VP
Joined: 22 Nov 2007
Posts: 1034

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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!
SVP
Joined: 29 Aug 2007
Posts: 2345

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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 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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GT

SVP
Joined: 29 Aug 2007
Posts: 2345

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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 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?
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

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CEO
Joined: 17 Nov 2007
Posts: 3420
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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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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SVP
Joined: 29 Aug 2007
Posts: 2345

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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
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me.
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Manager
Joined: 18 Jul 2009
Posts: 164
Location: India
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25 Jul 2009, 11:02
3
its 5C3 x 2C2 = 10 x 1 = 10
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Bhushan S.
If you like my post....Consider it for Kudos

Intern
Joined: 09 Sep 2009
Posts: 18

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24 Sep 2009, 20:50
1
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!)
Manager
Joined: 18 Jul 2009
Posts: 164
Location: India
Schools: South Asian B-schools

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26 Sep 2009, 13:41
1
consider placing 3 things in 5 options...the rest 2 blues will get settled automatically...
so 5C3 = 10 ways
_________________

Bhushan S.
If you like my post....Consider it for Kudos

Manager
Joined: 27 Oct 2008
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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
Intern
Joined: 15 Nov 2009
Posts: 28
Schools: Kelley

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 53738

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24 Dec 2009, 10:44
8
1
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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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
Re: Combinations   [#permalink] 31 Dec 2009, 10:12

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