Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 Nov 2011
Posts: 87

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
04 Apr 2016, 23:05
HI Bunnel
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?



Manager
Joined: 01 Nov 2016
Posts: 56
Concentration: Technology, Operations

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
27 Mar 2017, 20:51
What is the chance that four chosen have nobody married to each other? Well we want to find \(\frac{chance of success}{number of outcomes}\). Without using combitronics:
Chance of success is 12*10*8*6 Total outcomes are 12*11*9*8
\(\frac{12*10*8*6}{12*11*10*9}=\frac{8*6}{11*9}=\frac{16}{33}\)



Manager
Joined: 12 Nov 2016
Posts: 128
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39 GMAT 2: 650 Q47 V33

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
16 Dec 2017, 07:51
metallicafan wrote: +1 D
A faster way to solve it:
\(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\) Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach?



Retired Moderator
Joined: 25 Feb 2013
Posts: 1124
Location: India
GPA: 3.82

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
16 Dec 2017, 08:41
Erjan_S wrote: metallicafan wrote: +1 D
A faster way to solve it:
\(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\) Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach? Hi Erjan_SThere is nothing wrong with the approach. As the numbers are simple so you can simply count the possible and total outcomes and then keep on removing the pairs. There could be multiple ways to solve counting problems. P & C is preferred because you get the value through direct formula. in case of simple numbers you can manually count the number of possible ways or selections, but when the numbers or situation become a bit complex then you will have to rely upon P & C.



Director
Joined: 14 Dec 2017
Posts: 502
Location: India

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
11 Jun 2018, 02:36
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?
A. 1/33 B. 2/33 C. 1/3 D. 16/33 E. 11/12 Total # of ways to select 4 people out of 12 people = 12C4 = (12*11*10*9)/1*2*3*4) = 11 * 5 * 9 Favorable outcomes = # of ways to select 4 people out of 6 couples such that none of them are married to each other The First person can be chosen in 12 ways. The Second person can be chosen in 10 ways, since we will remove the spouse of first person selected & then choose. The Third person can be chosen in 8 ways, since we will remove the spouse of second person selected, as well & then choose. The Fourth person can be chosen in 6 ways, since we will remove the spouse of third person selected, as well & then choose. Now we need the # of ways of selection & not the arrangement of the persons, hence order does not matter. To account for this we divide by 4!, which is the # of possible arrangements of 4 persons. Hence Favorable outcomes = (12 * 10 * 8 * 6)/4! = 240 Required Probability = 240/(11 * 5 *9) = 16/33 Answer D. Thanks, GyM
_________________



Manager
Joined: 03 Apr 2018
Posts: 62
Location: United Arab Emirates
GPA: 4

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
20 Jun 2018, 23:13
1ST STEP: TOTAL NO OF WAYS OF SELECTING 4 PPL OUT OF 12 IS 12C4
THIS 12C4 WILL COME IN THE DENOMINATOR
NOW, NUMBER OF WAYS OF SELECTING 4 PEOPLE OUT OF 6 COUPLES IS 6C4 AND WE CAN CHOOSE EACH COUPLE IN 2 WAYS....HUSBAND AND WIFE
HENCE, 2*2*2*2 IS THE TOTAL NO OF WAYS OF CHOOSING 4 PEOPLE
SO, WE GET 6C4*2*2*2*2/12C4



Intern
Joined: 02 Jul 2017
Posts: 15
GMAT 1: 650 Q48 V30
GPA: 3.2

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
09 Mar 2019, 10:53
Dear Bunuel chetan2uWhat if we first select 1 individual from each set of couple and then out of those individuals, select 4 individuals. (2C1 * 2C1 * 2C1 * 2C1 * 2C1 * 2C1) * 6C4 = 960 Then, Total no. of ways to choose 4 people out of 12 is = 12C4 = 495 But turns out that this method is wrong as the probability comes out to be greater than 1 Can please help in pointing out why selecting 6 individuals from 6 couples first, is wrong?



Math Expert
Joined: 02 Aug 2009
Posts: 8602

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
09 Mar 2019, 19:00
j.shivank wrote: Dear Bunuel chetan2uWhat if we first select 1 individual from each set of couple and then out of those individuals, select 4 individuals. (2C1 * 2C1 * 2C1 * 2C1 * 2C1 * 2C1) * 6C4 = 960 Then, Total no. of ways to choose 4 people out of 12 is = 12C4 = 495 But turns out that this method is wrong as the probability comes out to be greater than 1 Can please help in pointing out why selecting 6 individuals from 6 couples first, is wrong? There are repetitions in this.. One example.. Say 12, 34,56,78,910,1112 are pair.. Select one out of each 1,3,5,7,9,11 choose ways of choosing 4: two ways would be 1,3,5,7 or 1,3,5,9 Now, choose one as 1,3,5,7,9,12 4: two ways would be 1,3,5,7 or 1,3,5,9 Not only these two, there will be many repetitions.. That is why you get probability more than 1.. But if you take 12 people first, remove the partner of the one you have chosen .. So 10 left, close one , and again remove the partner.. Now you are left with 8... And so on
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6286
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
23 Sep 2019, 10:56
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?
A. 1/33 B. 2/33 C. 1/3 D. 16/33 E. 11/12 total probability ; 12/12*10/11*8/10*6/9 = 16/33 IMO D



Yale Moderator
Joined: 05 May 2019
Posts: 181
GPA: 3

Re: If 4 people are selected from a group of 6 married couples
[#permalink]
Show Tags
25 Sep 2019, 03:17
There's an easy and efficient way to solve these kinds of Combination Questions.
First, You'll have to find the total possible ways to arrange 12 people into 4 places in the committee. This is without repetition, so the seats will be filled in the following ways.
\(12 , 11 , 10 , 9\) => So, a total of \(12*11*10*9\) ways.
Then, you're supposed to find the number of events where these four spots are filled in such a way that among the 12 people who are divided into 6 couples, each seat will be filled only by one representative from among the couples.
\(12 ,10 , 8 , 6\) => So, \(12*10*8*6\) ways.
To find the probability, \(\frac{12*10*8*6}{12*11*10*9}\)
Simplify and arrive at an answer of \(\frac{16}{33}\)




Re: If 4 people are selected from a group of 6 married couples
[#permalink]
25 Sep 2019, 03:17



Go to page
Previous
1 2
[ 30 posts ]

