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

 It is currently 25 May 2017, 01:51

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

What are the total number of ways, in which 3 red and 3 blue

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There
Followers: 1

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

What are the total number of ways, in which 3 red and 3 blue [#permalink]

Show Tags

12 May 2004, 18:53
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What are the total number of ways, in which 3 red and 3 blue beads can be put in a circular bracelet?
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6

Kudos [?]: 101 [0], given: 0

Show Tags

12 May 2004, 20:58
3 ways assuming all red beads are identical and all blue beads are identical

Following ways are possible
BBBRRR - You can rotate them to get different combinations
BBRBRR
BRBRBR
These 3 basics arrangements cover all possible arrangements
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4289
Followers: 43

Kudos [?]: 469 [0], given: 0

Show Tags

12 May 2004, 21:29
anandnk wrote:
3 ways assuming all red beads are identical and all blue beads are identical

Following ways are possible
BBBRRR - You can rotate them to get different combinations
BBRBRR
BRBRBR
These 3 basics arrangements cover all possible arrangements

I think believe my answer is wrong and 4 should be it by laying out the possible arrangements.
_________________

Best Regards,

Paul

SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6

Kudos [?]: 101 [0], given: 0

Show Tags

12 May 2004, 22:23
Yeah I agree I missed RBBRRB or BBRRBR

I dont know how to solve this problem using formulas.
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 17 [0], given: 0

Show Tags

12 May 2004, 22:44
anandnk wrote:
Yeah I agree I missed RBBRRB or BBRRBR

I dont know how to solve this problem using formulas.

Anand, as a programmer, you could write some algorithm to solve it. My friend solved this problem for N red and N green (it was given at all-Kazakh programming contest ). Kazakhstan is a large country in Central Asia (former Soviet Union).

4.

The method is as follows:

1. Think about combinations where all Bs are actually non-sequential. There is only 1 such comb.

2. Think about combs where only 2 Bs are sequential, but the remaining is not. => 2 combs.

3. Think about combs where all three B are sequential => there is only 1 such comb.

=> 4 is the answer.

The general method for (N greens, N reds) is as follows:

1. with no neighbors = 1.

2. with only 1 pair of 2 neighbors = N-1.

3. with only 2 pairs of 2 neighbors, ... etc.

So, it can be easily done even for N = 4,5.
Intern
Joined: 21 Mar 2004
Posts: 11
Location: Evansville, IN
Followers: 0

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

Show Tags

13 May 2004, 10:35
is 4 official answer?
I am not getting this . any help will be appreciated.
I got 8.
Any circular bracelet with six positions can be imagined as

_ _ _ _ _ _

we are required to sit 3 R balls and 3 B balls in six spaces.

so 2 X 2 X 2 X 1 X 1 X 1 = 8
_________________

Best wishes
Chetan

Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 17 [0], given: 0

Show Tags

13 May 2004, 10:49
sunny_god76 wrote:
is 4 official answer?
I am not getting this . any help will be appreciated.
I got 8.
Any circular bracelet with six positions can be imagined as

_ _ _ _ _ _

we are required to sit 3 R balls and 3 B balls in six spaces.

so 2 X 2 X 2 X 1 X 1 X 1 = 8

You got 8 because symmetric allocations were counted TWICE.

RBRBRB == BRBRBR! You did the job twice...

A propos, how did you get 2 X 2 X 2 X 1 X 1 X 1? It should rather be C[3,6] = 6!/(3!*3!) = 24... The fact that there are problems with symmetry makes this problem not so easy. As I've already said, my friend wrote a program to calculate total # of combinations in general case...
13 May 2004, 10:49
Display posts from previous: Sort by

What are the total number of ways, in which 3 red and 3 blue

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.