Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 23:40

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2741
Location: New York City
Followers: 11

Kudos [?]: 942 [0], given: 4

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

### Show Tags

25 Oct 2007, 10:20
12
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

64% (01:37) correct 36% (01:06) wrong based on 545 sessions

### HideShow timer Statistics

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, 16:14, edited 2 times in total.
Senior Manager
Joined: 04 Jun 2007
Posts: 367
Followers: 1

Kudos [?]: 75 [0], given: 0

### Show Tags

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: 2741
Location: New York City
Followers: 11

Kudos [?]: 942 [0], given: 4

### Show Tags

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: 466
Followers: 1

Kudos [?]: 49 [0], given: 1

### Show Tags

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: 100
Location: Astana
Followers: 1

Kudos [?]: 27 [0], given: 0

### Show Tags

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: 861
Followers: 17

Kudos [?]: 307 [1] , given: 0

### Show Tags

22 Dec 2007, 07:13
1
KUDOS
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
Followers: 0

Kudos [?]: 41 [0], given: 0

### Show Tags

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: 793
Followers: 2

Kudos [?]: 330 [0], given: 0

### Show Tags

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: 1083
Followers: 8

Kudos [?]: 567 [0], given: 0

### Show Tags

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: 2476
Followers: 70

Kudos [?]: 774 [0], given: 19

### Show Tags

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

_________________

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

GT

SVP
Joined: 29 Aug 2007
Posts: 2476
Followers: 70

Kudos [?]: 774 [0], given: 19

### Show Tags

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

GT

CEO
Joined: 17 Nov 2007
Posts: 3585
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3985 [0], given: 360

### Show Tags

19 Mar 2008, 22:17
Thu Oct 25, 2007 7:27 pm versus
Fri Oct 26, 2007 7:26 am

tricky GMAT TIGER
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

SVP
Joined: 29 Aug 2007
Posts: 2476
Followers: 70

Kudos [?]: 774 [0], given: 19

### Show Tags

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.
_________________

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

GT

Manager
Joined: 18 Jul 2009
Posts: 169
Location: India
Schools: South Asian B-schools
Followers: 2

Kudos [?]: 105 [3] , given: 37

### Show Tags

25 Jul 2009, 11:02
3
KUDOS
its 5C3 x 2C2 = 10 x 1 = 10
_________________

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

Intern
Joined: 09 Sep 2009
Posts: 19
Followers: 0

Kudos [?]: 13 [0], given: 2

### Show Tags

24 Sep 2009, 20:50
1
This post was
BOOKMARKED
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: 169
Location: India
Schools: South Asian B-schools
Followers: 2

Kudos [?]: 105 [1] , given: 37

### Show Tags

26 Sep 2009, 13:41
1
KUDOS
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
Posts: 185
Followers: 2

Kudos [?]: 150 [0], given: 3

### Show Tags

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: 29
Schools: Kelley
Followers: 0

Kudos [?]: 8 [0], given: 4

### Show Tags

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: 39042
Followers: 7751

Kudos [?]: 106491 [8] , given: 11626

### Show Tags

24 Dec 2009, 10:44
8
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________
Intern
Joined: 01 Oct 2009
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

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

Go to page    1   2    Next  [ 26 posts ]

Similar topics Replies Last post
Similar
Topics:
3 One red bottle and five blue bottles are to be arranged in a straight 8 11 Jan 2017, 11:10
2 Four red bottles and one blue bottle are to be arranged in a straight 1 23 Dec 2016, 04:15
1 Two red bottles and one blue bottle are to be arranged in a straight 1 13 Dec 2016, 08:08
18 A box contains 4 red chips and 2 blue chips. If two chips ar 10 24 Mar 2017, 09:25
11 There are 8 red chips and 2 blue ones. When arranged in a ro 13 18 Mar 2016, 21:06
Display posts from previous: Sort by