Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 14 Apr 2010
Posts: 124

If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
13 Aug 2010, 07:42
Question Stats:
70% (02:22) correct 30% (02:29) wrong based on 751 sessions
HideShow timer Statistics
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? A. 1/33 B. 2/33 C. 1/3 D. 16/33 E. 11/12
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 65807

Re: confuseddd
[#permalink]
Show Tags
13 Aug 2010, 08:10
bibha wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? 1/33 2/33 1/3 16/33 11/12 Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send only one "representatives" to the committee in \(C^4_6\) # of ways. But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\). So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\). Total # of ways to choose 4 people out of 12 is \(C^4_{12}\). \(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\) Answer: D. Similar problems with different approaches: combinationpermutationproblemcouples98533.html?hilit=married%20couplespscombinations94068.html?hilit=married%20couplescommitteeof88772.html?hilit=married%20couplesHope it helps.
_________________




Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 868
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs

If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
19 Feb 2012, 08:48
+1 D A faster way to solve it: \(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\)
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/myirlogbookdiary133264.html GMAT Club Premium Membership  big benefits and savings




Manager
Joined: 14 Apr 2010
Posts: 124

Re: confuseddd
[#permalink]
Show Tags
14 Aug 2010, 06:53
Why can't we do it like this:
total ways of selective 4 ppl from 6 married couples = 12C4 Favorable outcome = 12 *10*8*6 ????



Math Expert
Joined: 02 Sep 2009
Posts: 65807

Re: confuseddd
[#permalink]
Show Tags
14 Aug 2010, 07:27
bibha wrote: Why can't we do it like this:
total ways of selective 4 ppl from 6 married couples = 12C4 Favorable outcome = 12 *10*8*6 ???? The way you are doing is wrong because 12*10*8*6=5760 will contain duplication and if you are doing this way then to get rid of them you should divide this number by the factorial of the # of people  4! > \(\frac{5760}{4!}=240=C^2_4*2^8=favorable \ outcomes\). Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible: \(A_1,B_1\); \(A_1,B_2\); \(A_2,B_1\); \(A_2,B_2\). Only 4 such committees are possible. If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! > 8/2!=4. Hope it helps.
_________________



Intern
Joined: 29 Dec 2009
Posts: 18

Re: confuseddd
[#permalink]
Show Tags
16 Aug 2010, 11:36
awesome explanation +1



Intern
Joined: 04 Aug 2010
Posts: 20

Re: confuseddd
[#permalink]
Show Tags
13 Sep 2010, 19:38
But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2
Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain



Math Expert
Joined: 02 Sep 2009
Posts: 65807

Re: confuseddd
[#permalink]
Show Tags
13 Sep 2010, 20:20
harithakishore wrote: But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2
Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain We have 6 couples: \(A (a_1, a_2)\); \(B (b_1, b_2)\); \(C (c_1, c_2)\); \(D (d_1, d_2)\); \(E (e_1, e_2)\); \(F (f_1, f_2)\); We should choose 4 people so that none of them will be married to each other. The above means that 4 chosen people will be from 4 different couples, for example from A, B, C, D or from A, D, E, F... The # of ways to choose from which 4 couples these 4 people will be is \(C^4_6=15\); Let's consider one particular group of 4 couples: {A, B, C, D}. Now, from couple A in the group could be either \(a_1\) or \(a_2\), from couple B in the group could be either \(b_1\) or \(b_2\), from couple C in the group could be either \(c_1\) or \(c_2\), and from couple D in the group could be either \(d_1\) or \(d_2\). So each couple has two options (each couple can be represented in the group of 4 people by \(x_1\) or \(x_2\)), so one particular group of 4 couples {A, B, C, D} can give us \(2*2*2*2=2^4\) groups of 4 people from different couples. One particular group of 4 couples {A, B, C, D} gives \(2^4\) groups of 4 people from different couples > 15 groups give \(15*2^4\) groups of 4 people from different couples (total # of ways to choose 4 people so that no two will be from the same couple) . Hope it's clear.
_________________



Intern
Joined: 04 Aug 2010
Posts: 20

Re: confuseddd
[#permalink]
Show Tags
13 Sep 2010, 20:27
Thats a fantabulous explanation...thankyou so much....



Intern
Joined: 19 Dec 2011
Posts: 4

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
19 Jan 2012, 00:13
Total possible selection = 12!/(4!*8!)= 11*45 (after simplification) Favourble out come can be obtained by the multiplying the following combinations. 1. We require only 4 people. So these 4 are going to be from 4 different groups. Total availbale grops =6. So this combination is 6c4 = 6!/(4!*2!) =15 2. Select 1 member from each group = 2c1*2c1*2c1*2c1=2^4=16
Probability = (15*16)/(11*45)=16/33



Senior Manager
Joined: 13 Aug 2012
Posts: 385
Concentration: Marketing, Finance
GPA: 3.23

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
27 Dec 2012, 18:34
bibha wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? 1/33 2/33 1/3 16/33 11/12 If we are to select 4 people from 6 couples WITHOUT any restriction, how many ways can we make the selection? 12!/4!6! = 11*5*9 = 495 If we are to select 4 people from 6 couples WITH restriction that no married couple can both make it to the group, only a representative? 6!/4!2! = 15 But we know that to select a person from each couple, take 2 possibilities15*2*2*2*2 = 240 Probability = Desired/All Possibilities = 240/495 = 16/33 Answer: D



Manager
Joined: 15 Aug 2013
Posts: 223

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
23 Apr 2014, 19:42
metallicafan wrote: +1 D
A faster way to solve it:
\(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\) I believe that I had seen elsewhere that IF we were doing this same problem without the probability part of the question, we would have to divide (12x10x8x6) with 4!. Why is that not applicable when doing probability? Don't we still need the favorable outcomes?



Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
WE: Analyst (Consulting)

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
23 Apr 2014, 22:28
Number of ways to select...Atleast one married couple is
6C1 Choose 1 married couple for 2 seats 10C2 Chose 2 members from the remaining 5 couples for the remaining 2 seats
6C1*10C2= 270 No of ways to select 4 people from 12=12C4
P= 270/495
Answer=1 the prob of the above =5/11
What am I missing here?



Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
WE: Analyst (Consulting)

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
Updated on: 24 Apr 2014, 22:34
Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions?
Originally posted by JusTLucK04 on 23 Apr 2014, 22:49.
Last edited by JusTLucK04 on 24 Apr 2014, 22:34, edited 1 time in total.



Manager
Joined: 15 Aug 2013
Posts: 223

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
24 Apr 2014, 18:20
JusTLucK04 wrote: Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions? I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10?



Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
WE: Analyst (Consulting)

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
24 Apr 2014, 23:02
russ9 wrote: JusTLucK04 wrote: Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions? I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10? I think it should be...100*99....91*90 And if the question mentions that it is possible that a student recieves not even a single pencil..I think we go case wise with 1 student gets all..2 student get all pencils..and so on



Intern
Joined: 24 Apr 2014
Posts: 8

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
26 Apr 2014, 01:38
1. Select the 1st person: sure there is 1st'spouse in the group > then 11 left 2. Select the 2nd person: probability not choose 2rd's spouse is 10/11 > then 10 left (2 ppl are 1st and 2nd's spouses) 3. Select the 3rd person: probability 8/10 > then 9 left (3ppl are 1st, 2nd and 3rd spouses) 4. Select the 4th: probability 6/9 > Probability when choose 4 ppl = 10/11*8/10*6/9 = 11/33



Manager
Joined: 27 May 2014
Posts: 74

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
27 Jun 2014, 16:44
Bunuel  what's wrong with the following approach:
6c2  two couples 15 6c1 * 10c2  270
1(270/495+15/495) = 210/495= 14/33??
Posted from my mobile device



Math Expert
Joined: 02 Sep 2009
Posts: 65807

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
28 Jun 2014, 04:04
bankerboy30 wrote: Bunuel  what's wrong with the following approach:
6c2  two couples 15 6c1 * 10c2  270
1(270/495+15/495) = 210/495= 14/33??
Posted from my mobile device 10C2 can also give second couple which is already counted from 6C2.
_________________



Intern
Joined: 13 Oct 2014
Posts: 1
Concentration: Marketing, General Management
GPA: 3.5
WE: Sales (Pharmaceuticals and Biotech)

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
15 Dec 2014, 06:21
Bunnel, Whats wrong with this approach ? Case1 12c46c2 ( Both couples selected) Case2 12c4 6c1 * 10c1 * 8c1( One couple & 2 Noncouple )
Sum up the putcomes of both the cases & find probability.
Please tell how to approach this




Re: If 4 people are selected from a group of 6 married couples
[#permalink]
15 Dec 2014, 06:21



Go to page
1 2
Next
[ 30 posts ]

