Given that there are 6 married couples. If we select only 4

Question

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

A)  \frac{1}{3}

B)  \frac{1}{8}

C)  \frac{16}{33}

D)  \frac{2}{3}

E)  \frac{288}{495}

walker · Accepted Answer

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

or

p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}

or

p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}

walker · Answer

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

\frac{12}{12}  - we choose 12 of 12.

\frac{10}{11}  - we choose 10=12-1(prevous choice)-1(another people out of couple) of 11=12-1(prevous choice).

\frac{8}{10}  - we choose 8=12-2(prevous choice)-2(another people out of couple) of 10=12-2(prevous choice).

\frac{6}{9}  - we choose 6=12-3(prevous choice)-3(another people out of couple) of 9=12-3(prevous choice).

_______________________

p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33} 
the same logic as for  C_m^n

GMATBLACKBELT · Answer

Prob: 1*10/11*8/10*6/9 = 16/33

walker · Answer

C^6_4  - we choose 4 couples of 6 ones.

C^2_1  - we chose one people of 2 ones for one couple.

(C^2_1)^4  - we have 4 couple and for each we choose one people of 2 ones for one couple.

C^{12}_4  - the total number of combinations to choose 4 people from 12 people.

x2suresh · Answer

= (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4
 = 12*10*8*6 / (12*11*10*9) = 16/33

walker · Answer

C^6_4  - we choose 4 couples (not people) out of 6 couples (the number of all couples)

the next step: we choose one person out of each couple -  C^2_1

xALIx · Answer

Walker, 
Ok, its making better sense now. 
Are we choosing 4 couples because that means "not 4 people", ie 12-4=8?

walker · Answer

So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.
There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple -  C^6_4*(C^2_1)^4

look at other problem:

we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

I hope it is clearer now.

tusharvk · Answer

total ways of choosing 4 out of 12:12x11x10x9
first person can be chosen in 12 ways. The next person can be chosen in 10 ways (because we don't want spouses to be in the group). The next person can be again chosen in only 8 ways (out of the 10 people left, we have to exclude 2 whose spouses we have already selected). And the last person in 6 ways.

hence, probability=12x10x8x6/(12x11x10x9) = 16/33.

x2suresh · Answer

12C1 = select any person from 6 married couples (12 person)
10C1 = select second person from remaining people and exclude the first person's spouse
8C1 = select 3rd person from remain people exclue first and second perssons's spouses
6C1 = select 4th person from remaining people exclude 1st,2nd ,3rd persons's spouse

Becuase order is not matter.. you need to divide by 4!

Did you get it?

=

sanjay_gmat · Answer

another way of looking at this problem :

total number of ways = 12C4 = 495

now let's try to find out total number of unfavourable ways. We can subtract those from 495 to get the number of favourable ways.

number of ways of choosing 4 people such that there are two couples = 6C2 = 15

number of ways of choosing 4 people such that there is only one couple = number of ways of choosing one couple * number of ways of choosing two people who are not couples = 6C1 (10C2 -5) = 240.

total number of unfavourable ways = 240 + 15 = 255

total number of favourable ways = 495 - 255 = 240

required probability = 240/495 = 16/33

srivas · Answer

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

Soln:
= (12/12) * (10/11) * (8/10) * (6/9)
= 16/33

Bunuel · Answer

bumpbot · Answer

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