Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
Intern
Joined: 10 Oct 2003
Posts: 41
Location: Finland

Given 10 beads on a necklace, 6 white and 4 red, how many
[#permalink]
Show Tags
25 Nov 2003, 02:43
Given 10 beads on a necklace, 6 white and 4 red, how many ways
canthe beads be arranged so that no three red beads are together? == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



VP
Joined: 03 Feb 2003
Posts: 1400

in my opinion:
total ways = 10!
forbidden ways = unite three red balls in one black: 4C3 ways=4
thus, we have 1 black, 1 red, and 6 white balls; there are 8!/6!*1!*1! ways to do so; or 7*8=56
10!4*56=10!224



Intern
Joined: 20 Oct 2003
Posts: 13
Location: Moscow

Let's calculate cases that undergo the limitation: let's assume that 1st, 2nd and 3rd places are taken by red bean. Then if either 4th or 10th place is taken by red bean as well would mean that 4 red beans are together  which is to me also under the limitation. Thus I would solve it this way:
total = 10!
possible combinations of red beans in a row = 4!
possible combinations of white beans when 4 red are in a row = 6!
outcome:
10!  (6!*4!)



GMAT Instructor
Joined: 07 Jul 2003
Posts: 654
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

stolyar wrote: in my opinion:
total ways = 10! forbidden ways = unite three red balls in one black: 4C3 ways=4 thus, we have 1 black, 1 red, and 6 white balls; there are 8!/6!*1!*1! ways to do so; or 7*8=56
10!4*56=10!224
be careful. this problem has a lot of hidden complexities if properly solved. 1) a necklace can be rotated such that a number of arrangements are henceforth "indistinguishable". 2) a necklace with beads can also be "flipped over" (mirror image) adding yet another "axis of symmetry"  these are hard to count because we need to exclude arrangement that are already symmetrical in order to avoid double counting.
Consider this: suppose we have a bracelet with 3 different colored beads (assume that there is no distiguishing clasp and the beads are symmetricallyl distributed around the bracelet). There are theoretically 3! or 6 different ways to arrange the beads. Suppose you had six bracelets each with one of those arrangements, then threw them all into a hat and mixed them up. Now dump them out of the hat and try to distinguish one for the other. You will realize that they are ALL the same arrangement!
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



Senior Manager
Joined: 14 Oct 2003
Posts: 490
Location: On Vacation at My Crawford, Texas Ranch

Akamai,
Side question; Isn't the total number of arrangements of objects (beads or whatever) on a circular ring calculated by the formula (n1)! ? Thanks.



VP
Joined: 03 Feb 2003
Posts: 1400

I assumed that a necklace is a string, not a circle.



GMAT Instructor
Joined: 07 Jul 2003
Posts: 654
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

stolyar wrote: I assumed that a necklace is a string, not a circle.
I don't know how reasonable that assumption is (never seen anyone wear a necklace as a string) but even if that were true, you still missed the mirror image symmetry.
_________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



SVP
Joined: 30 Oct 2003
Posts: 1509
Location: NewJersey USA

I thought the number of combinations are ( assuming the necklace is a string instead of a circular necklace )
[ 10! / ( 6! * 4! ) ]  4 * [ 8! / 6! * 1! * 1! ]
Can anyone explain me why this formula is wrong ( if it is wrong )



Intern
Joined: 04 Oct 2003
Posts: 41
Location: Silicon Valley, CA

I agree with anandnk. But have a small addition
[10!/(6!*4!)]  4*[8!/(6!*2!)]
The reason for the extra 2! is that even though we take 3 red together, we will have 1 more red bead. And the (group or 3 red) + 1 more read make 2 simillar items.
Take also makes me think that we might not have counted the cases correctly.
Consider this: Rg Rg Rg are the three Red beads taken together. &
Ra is the 4th red bead.
say two Rg's are together (Rg, Rg) and Ra is next to it, this will make 3 red together, though we don't have all Rg's toether. Reason is that we have 4 reds and we are considering only 3 together and the 4th is in don't care position. I feel we can't do this. The 4th red bead also has to be accounted for.
OR may be my head is twisted too bad. == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



NonHuman User
Joined: 09 Sep 2013
Posts: 13242

Re: Given 10 beads on a necklace, 6 white and 4 red, how many
[#permalink]
Show Tags
26 Sep 2018, 06:38
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Given 10 beads on a necklace, 6 white and 4 red, how many
[#permalink]
26 Sep 2018, 06:38






