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

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

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New post 05 Apr 2016, 00:05
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, 21:51
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, 08:51
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, 09:41
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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, 03:36
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,
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 21 Jun 2018, 00:13
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
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 09 Mar 2019, 11:53
Dear Bunuel chetan2u

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

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New post 09 Mar 2019, 20:00
1
j.shivank wrote:
Dear Bunuel chetan2u

What 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 1-2, 3-4,5-6,7-8,9-10,11-12 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
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 23 Sep 2019, 11:56
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
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Re: If 4 people are selected from a group of 6 married couples  [#permalink]

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New post 25 Sep 2019, 04:17
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 \(\frac{16}{33}\)
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Re: If 4 people are selected from a group of 6 married couples   [#permalink] 25 Sep 2019, 04:17

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