Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 02 Nov 2009
Posts: 20

A committee that includes 6 members is about to be divided [#permalink]
Show Tags
21 Apr 2010, 01:10
3
This post received KUDOS
15
This post was BOOKMARKED
Question Stats:
47% (01:05) correct 53% (01:23) wrong based on 271 sessions
HideShow timer Statistics
A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? A. 10% B. 20% C. 25% D. 40% E. 50%
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by reto on 02 Sep 2015, 04:51, edited 1 time in total.
official answers, source!



Manager
Joined: 18 Mar 2010
Posts: 86
Location: United States

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
21 Apr 2010, 10:25
4
This post received KUDOS
2
This post was BOOKMARKED
Committee #1: 50% chance that David will be a member, leaving 2 possible spots the other 5 people. Michael's chances of being one of those 2 is 2/5 or 40%. Chances of David and Michael being on this committee is 50%x40%=20%.
Committee #2: Same.
20% + 20% = 40%.



Manager
Joined: 05 Mar 2010
Posts: 192

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
21 Apr 2010, 23:38
1
This post received KUDOS
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? Here if we first calculate total number of subcommittes possible for 1st subcommittee  selection of any 3 out of 6 = 6C3 = 20 fOR 2nd sub committee  3 out of remaining 3 = 1 total subcommittees possible are 20+1 = 21 Now if we take michael and david as person (so both will be in one committee) total possible committee possible are 5C3 = 10 10 is definately not 40% of 21 Where am i going wrong? sometimes i really get confused !!!!
_________________
Success is my Destiny



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2699
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
22 Apr 2010, 11:46
8
This post received KUDOS
1
This post was BOOKMARKED
A / B <==> _ _ _ / _ _ _ where A and B are two committees. For A, 2 places are filled by David and Michel, third can be select in 4c1 ways. For B , 3 out of remaining three can be selected in 3c3 total required ways = 4c1*3c3*2 ( we have multiplied with 2 , as they can be part of either 2) To calculate total ways : For A , 3 places can be filled by 6c3 and remaining by 3c3 , here we wont multiply by 2, as there is no specification. for B, 3 out of remaining three can be selected in 3c3 Now probability = \(\frac{(total required ways)}{(total ways)}\) = \(\frac{4c1*3c3*2}{6c3*3c3}\) = \(\frac{4*2}{6c3}\) = \(\frac{4*2*3!*3!}{6!}\) = \(\frac{4*2*3!}{6*5*4}\) = \(\frac{2}{5}\) 2/5 = 2*100/5 % = 40%
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 05 Mar 2010
Posts: 192

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
23 Apr 2010, 00:32
3
This post received KUDOS
i got it now, even if we assume both david and michael as one person we are left with 4 and now we have to choose 1 out of 3 and for 2nd committee remaining 3 out of 3. Because david and michael can be in any of the committe so multiply t by 2 As i said sometimes i get confused in simple solution. i should do quant with fresh mind not when i am tired
_________________
Success is my Destiny



Senior Manager
Joined: 27 May 2012
Posts: 459

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Sep 2012, 00:51
2
This post received KUDOS
1
This post was BOOKMARKED
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? I found another method more helpful and understandable but missing from this topic hence contributing . Hope it helps total number of 3 person , 2 sub committees \(\frac{C^6_3 * C^3_3}{2!}\) = 10 ( we divide by 2! as the two subcommittees order is not important we do not have \(1^{st}\) subcommittee and \(2^{nd}\) sub committee. Bunuel can enlighten further for beginners in this area ). now suppose M and D are members of a sub committee then the remaining person can be chosen in \(c^4_1\) way , so \(C^1_1 * C^1_1 *C^4_1\)= 4 total ways a sub committee can form with M and D as members hence percentage \(\frac{4}{10} *100\) = 40%
_________________
 Stne



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Sep 2012, 05:54
2
This post received KUDOS
2
This post was BOOKMARKED
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? Count how many subcommittees Michael is a member of: M * *  means Michael plus two other persons; for the two stars we can choose from 5 members, therefore 5C2=5*4/2=10 different subcommittees. Count how many subcommittees both Michael and David are members of: M D *  means both Michael and David are in the same subcommittee, plus one extra person: for the star we can choose from the 4 remaining members, therefore 4 different subcommittees. This gives 4/10 = 40%.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Intern
Joined: 14 Jul 2012
Posts: 3

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Sep 2012, 06:09
6
This post received KUDOS
1
This post was BOOKMARKED
This is equivalent to: We have 6 chairs divided in 2 Groups: A and B each with 3 chairs. If Michael sits on the first chair in the group A, what is the posibility that David sits on any of another 2 available chairs in Group A?
So: A: M _ _ / B: _ _ _ There are 2 chairs in Group A for favorable outcomes and 5 available chairs as total possible outcomes. The answer is 2/5



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Sep 2012, 06:52
sstasic wrote: This is equivalent to: We have 6 chairs divided in 2 Groups: A and B each with 3 chairs. If Michael sits on the first chair in the group A, what is the posibility that David sits on any of another 2 available chairs in Group A?
So: A: M _ _ / B: _ _ _ There are 2 chairs in Group A for favorable outcomes and 5 available chairs as total possible outcomes. The answer is 2/5 That's an excellent reinterpretation of the question! How did you come up with it?
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 24 Jul 2011
Posts: 75
Location: India
Concentration: Strategy, General Management
WE: Asset Management (Manufacturing)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Sep 2012, 07:37
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? There are 6 members, whom we need to divide into 2 sub committees. So, in each committee we'll have 3 members. If we form committees with the help of combination, we'll get \(^6C 3\)\(= 20\) number of committees. Now, let's see how we'll get 10 numbers of unique committees assume we have following members. a, b, c, x, y, and z. The one of committees will be formed by choosing a, b and c. Naturally x,y and z are left out. Similarly, when we choose x,y and z to form a committee a,b and c are left out. This means unique number of committees \(= 20/2\)\(=10\) Now, let's find the committees in which M and D are members If M and D are already selected we need to select one more member out of 4 members. This means we can select another member in 4 ways. So, % of committees in which both M and D are present \(= 4/10\) \(= 40 %\)
_________________
My mantra for cracking GMAT: Everyone has inborn talent, however those who complement it with hard work we call them 'talented'.
+1 Kudos = Thank You Dear Are you saying thank you?



Math Expert
Joined: 02 Sep 2009
Posts: 44373

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
11 Sep 2012, 03:13



Senior Manager
Joined: 03 Sep 2012
Posts: 392
Location: United States
Concentration: Healthcare, Strategy
GPA: 3.88
WE: Medicine and Health (Health Care)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
25 Sep 2012, 06:04
1
This post received KUDOS
Total number of Members 6 Size of the new sub committees. 3 x2 ( two sub committees)  Let these be Committee A and B Chances of Micheal being in Comm. A = 1/2 Total no of spots left in comm A = 2 , Members that can possibly take these spots = 5 ( Micheal ie. the sixth member has all ready taken his spot in comm. A ) therefore David has 2/5 chance of being in committee A ... Multiplying the two (to figure out the chances of both of them being in the same committee at the same time) we get ( 2/5 ) x 1/2 = 1/5 or 20 % ... Therefore , Chances that both Micheal and David will be together in the FIRST committee are 20% . They have equal chance ie. 20% of being together in Sub Comm. B (instead of A) therefore the Answer is 20x2 = 40 % ....
_________________
"When you want to succeed as bad as you want to breathe, then you’ll be successful.”  Eric Thomas



Intern
Joined: 14 Jul 2012
Posts: 3

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
10 Oct 2012, 04:51
1
This post received KUDOS
In school i was taught to resolve problems in the form of A+B=?. It become much more interesting when i found in real life practical uses for theoretical concepts (especially through computer programming of industrial machinery to obtain a specific real product). The Gmat approach of math is similar to real life one: it takes a life experience and expects you to resolve it. It's like driving a car and somebody asks how much more will we be driving. There is no given equation. To give an answer you first try to find what elements you need: you look at your driving speed and try to get an idea of the distance left. You might have a GPS and get the distance there, you might know the total trip length and setted a trip counter on your board when you left your home. You take the easiest information you have. Maybe you wait a few moments to get in front of a board with distance info and you might have there your destination. Then you use theory concept that Time=Distance / Speed, use the actual numbers and solve the question. Gmat is placing a questions and lets you choose from theoretical concepts you need to answer it. Since every math theorem is demonstrated using simpler previous theorems, it means that you can solve a problem in the most rudimental way (by counting each possible outcome in a probability question) or use derivative concepts for general outcome (combinations for example). There is a multitude of ways to resolve a problem; some of them are faster than others. Sometime counting each possible outcome is faster than using general formulas. The idea is to see what the equation is and if it can be simplified. Maybe 1 (probability for wrong outcomes) is easier to calculate in probability question. When you take the approach of calculating the opposite numbers, you are actually rephrasing the question. The equation would be the same for various other life examples, so try to rephrase it. Take if the current question: ”On what percent of the possible subcommittees that Michael is a member of is David also a member?”. The part with “on what percent of the possible subcommittees” is equivalent to “what is the probability”. The other part just asks David to be in the same Committee with Michael. So: “what is the probability that Michal is in the same committee as David”. The outcome will be the same if you change the layout of the problem. Let’s say that there is a club with 6 members and there should be formed a committee of 3 members: president, treasurer and secretary. Michael is president. What is the probability that David would be in the committee (either as secretary or treasurer)? The information that the rest of the members form or do not form a committee (for a given project for example) is useless. The actual question comes to the probability of occupying one of the 2 given vacant positions in the committee by a specific member from other 5 members of the club! :D (< this is the easier form of the problem I can get to). EvaJager wrote: sstasic wrote: This is equivalent to: ... That's an excellent reinterpretation of the question! How did you come up with it?



Intern
Joined: 24 Oct 2012
Posts: 3
WE: Business Development (Consulting)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
25 Oct 2012, 21:52
Let the committee members be A,B,C,D,E and F, where A and B corresponds to Michael and David. The possible combinations in which 1st three committee member can be formed are  ABC ABD ABE ABF ACD ACE ACF ADE ADF AEF
Rest of the combinations would form a pair to one of the above mentioned combination. i.e, they would form the 2nd committee of one of the above.
Total combinations in which A and B are together = 4 Total number of combinations = 10
So percentage in which A and B are together = 4/10 = 2/5.
May be this method takes a bit longer, but I guess its still helpful in understanding the concept and will avoid confusion.



Intern
Joined: 02 Sep 2010
Posts: 43
Location: India

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
28 Oct 2012, 05:39
gurpreetsingh wrote: A / B <==> _ _ _ / _ _ _ where A and B are two committees.
For A, 2 places are filled by David and Michel, third can be select in 4c1 ways.
For B , 3 out of remaining three can be selected in 3c3
total required ways = 4c1*3c3*2 ( we have multiplied with 2 , as they can be part of either 2)
To calculate total ways :
For A , 3 places can be filled by 6c3 and remaining by 3c3 , here we wont multiply by 2, as there is no specification.
for B, 3 out of remaining three can be selected in 3c3
Now probability = \(\frac{(total required ways)}{(total ways)}\)
= \(\frac{4c1*3c3*2}{6c3*3c3}\) = \(\frac{4*2}{6c3}\) = \(\frac{4*2*3!*3!}{6!}\) = \(\frac{4*2*3!}{6*5*4}\) = \(\frac{2}{5}\)
2/5 = 2*100/5 % = 40% Hey Gurpreet, Can you please have a look at the below question. The answer should be 3/5 according to your approach. acommitteeof3hastobeformedrandomlyfromagroupof81051.htmlOnly thing I am confused about is, if we actually need to multiply the numerator by 2. In this question you have done that, if that is correct way of doing it, why is that multiplication is not required in the below question. Bunnel, can you please help too? Thanks
_________________
The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2699
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
28 Oct 2012, 07:13
hermit84 wrote: Hey Gurpreet, Can you please have a look at the below question. The answer should be 3/5 according to your approach. acommitteeof3hastobeformedrandomlyfromagroupof81051.htmlOnly thing I am confused about is, if we actually need to multiply the numerator by 2. In this question you have done that, if that is correct way of doing it, why is that multiplication is not required in the below question. Bunnel, can you please help too? Thanks For my solution I assumed A and B committee to be specific. So I multiplied by 2. If you do not want to do so..you can leave it to 4c3 but while calculating you have to divide 6c3 by 2! to remove the duplicate values. again the value would be same..as we will divide by 2! only when we do not care about the committee being specific.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



SVP
Joined: 14 Apr 2009
Posts: 2146
Location: New York, NY

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
30 Oct 2012, 14:48
Just answered a similar question:
Here's some clarification for those who are still confused.
First group of 6
     
Gets broken into 2 subcommittees of 3 each:
     
We know Michael is already in one of them:
M     
We just need to fill in one of those two slots next to Michael with an "A" for Anthony.
So we already know Mike is in that slot and that there are 5 remaining choices.
Well, how many ways can we pick Anthony such that he ends up in Michael's group?
Keep in mind that order matters so we can multiply a line of nCr formulas:
[ (Out of 1 available Anthony, pick that 1 Anthony for that 2nd spot) * (Out of the remaining 4, choose any 1 for that 3rd spot) ] =  (Out of the initial 5 remaining people, choose 2 to fill up the 2nd and 3rd slots)
= [ (1C1) * (4C1) ] / (5C2)
= 4 / 10
= 40%
Hope that helps.



Retired Moderator
Joined: 29 Apr 2015
Posts: 883
Location: Switzerland
Concentration: Economics, Finance
WE: Asset Management (Investment Banking)

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
02 Sep 2015, 04:52
bumpbot wrote: 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. Bumping up for Review, after adding official AC & source
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.



Manager
Joined: 01 Jun 2013
Posts: 124

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
07 Sep 2015, 12:33
IMO, this is not a gmat type question, at least from wording perspective... It's not clear whether the M&D among 6. It was suppose to define that unambiguously. I thought that both are among 6, and if one is in subcategory, the chance would be in the same 50%.
_________________
Please kindly click on "+1 Kudos", if you think my post is useful



Manager
Joined: 14 Jul 2014
Posts: 190
Location: United States
GMAT 1: 600 Q48 V27 GMAT 2: 720 Q50 V37
GPA: 3.2

Re: A committee that includes 6 members is about to be divided [#permalink]
Show Tags
19 Jan 2016, 11:16
If first team has Michael and David, there will be 4 ways to arrange that team. M, D, _ . (Last member can be one of the remaining four). Second team will be arranged only in one way, order does not matter. So 4 * 1 = 4 ways for first team having M and D.
Repeat for second team = 4 ways.
Total number of possibilities: 6C3 = 20.
So, 8 possible ways out of 20 = 40 percent.




Re: A committee that includes 6 members is about to be divided
[#permalink]
19 Jan 2016, 11:16



Go to page
1 2
Next
[ 23 posts ]



