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:
combination-permutation-problem-couples-98533.html?hilit=married%20couples
ps-combinations-94068.html?hilit=married%20couples
committee-of-88772.html?hilit=married%20couples

Hope it helps.
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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
????

awesome explanation +1

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.
_________________

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

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?

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 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

Bunuel - what's wrong with the following approach:

6c2 - two couples 15
6c1 * 10c2 - 270

1-(270/495+15/495) = 210/495= 14/33??

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Bunnel,
Whats wrong with this approach ?
Case-1
12c4-6c2 ( Both couples selected)
Case-2
12c4- 6c1 * 10c1 * 8c1( One couple & 2 Non-couple )

Sum up the putcomes of both the cases & find probability.

Please tell how to approach this