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

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.

awesome explanation +1

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.

Thats a fantabulous explanation...thankyou so much....

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

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 possibilities
15*2*2*2*2 = 240

Probability = Desired/All Possibilities = 240/495 = 16/33

Answer: D

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.

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?

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

\(\frac{12*10*8*6}{12*11*10*9}=\frac{8*6}{11*9}=\frac{16}{33}\)

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

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

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 \(\frac{16}{33}\)

­Hello Bunuel - How I approached this is:

Case 1: Both married couples: 6c1*5c1 = 30
Case 2: 1 married couple: 6c1*[(10c1*8c1)/2!] = 240

Required case of no married couple = 12c4-(30+240) = 225
Required prob = 225/495 = 5/11

Where am I faltering Bunuel ?­