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

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?

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?

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 would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10?

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

1. Select the 1st person: sure there is 1st'spouse in the group --> then 11 left
2. Select the 2nd person: probability not choose 2rd's spouse is 10/11 --> then 10 left (2 ppl are 1st and 2nd's spouses)
3. Select the 3rd person: probability 8/10 --> then 9 left (3ppl are 1st, 2nd and 3rd spouses)
4. Select the 4th: probability 6/9
--> Probability when choose 4 ppl = 10/11*8/10*6/9 = 11/33

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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10C2 can also give second couple which is already counted from 6C2.

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