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If 4 people are selected from a group of 6 married couples

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Manager
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 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?
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 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

\(\frac{12*10*8*6}{12*11*10*9}=\frac{8*6}{11*9}=\frac{16}{33}\)
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 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?
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 16 Dec 2017, 08:41
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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_S

There 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.
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 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
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 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
GMAT Club Bot
Re: If 4 people are selected from a group of 6 married couples &nbs [#permalink] 20 Jun 2018, 23:13

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