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6 married couple are present at a party. If 4 people are

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Senior Manager
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6 married couple are present at a party. If 4 people are [#permalink] New post 07 Aug 2007, 19:23
6 married couple are present at a party. If 4 people are selected out of these 12, what is the probability that none of these people will be married to each other?

A) 1/33

B) 2/33

C) 1/3

D) 16/33

E) 11/12
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 [#permalink] New post 07 Aug 2007, 22:43
xsaudx wrote:
the answer is A


I know the answer. It's definitely not A. Would you mind sharing your thought process?
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 [#permalink] New post 07 Aug 2007, 23:02
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

its kind of tricky but think about this a bit and you will understand.
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 [#permalink] New post 07 Aug 2007, 23:18
empty_spaces wrote:
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

its kind of tricky but think about this a bit and you will understand.


Thanks. This is the right answer.

By the way, I was trying to solve it in the following manner:

number of ways of selecting 4 people out of 12 = 12C4 = 495.

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

This is greater than 495. What is my mistake?

I also tried, finding the number of ways we can pick couples among these 4 people.

First person can be chosen in 12 ways.
Second person can be chosen in only 1 ways.

third person can be chosen in 10 ways.
fourth person can be chosen in 1 way

so the total number of favourable ways = 495 -120=375

P=375/495.

could you please point out my errors?
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 [#permalink] New post 07 Aug 2007, 23:54
Let me try :


now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.


above statement is wrong in this context ....you are mixing permutations and combinations .

if you want solve this way then divide "12*10*8*6" by 12P4 instead of 12C4..and you should have the correct answer.

yet another method :

consider each couple as one and then find : 6C4 for selecting 4 out of 6 couples but now since you only want one out of each of these 4 couples you need to multiply this by 2 for each couple which gives you ...2 * 2 * 2 * 2* 6C4

now your answer is : (2 * 2 * 2 * 2* 6C4 ) / 12 C 4 ...solving that should give you the correct answer





shoonya wrote:
empty_spaces wrote:
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

its kind of tricky but think about this a bit and you will understand.


Thanks. This is the right answer.

By the way, I was trying to solve it in the following manner:

number of ways of selecting 4 people out of 12 = 12C4 = 495.

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

This is greater than 495. What is my mistake?

I also tried, finding the number of ways we can pick couples among these 4 people.

First person can be chosen in 12 ways.
Second person can be chosen in only 1 ways.

third person can be chosen in 10 ways.
fourth person can be chosen in 1 way

so the total number of favourable ways = 495 -120=375

P=375/495.

could you please point out my errors?
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 [#permalink] New post 08 Aug 2007, 01:28
D

P (1st Person) = 12/12 - We can choose anybody
P (2nd Person) = 10/11 - Out of remaining 11 we can choose 10 because we can not choose who is already selected and his/her spouse.
P (3rd Person) = 8/10 - Out of remaining 10 we can choose 8 because we can not choose who are already selected (2 selected) and their 2 spouses.
P (4th Person) = 6/9 - Out of remaining 9 we can choose 6 because we can not choose who are already selected (3 selected) and their 3 spouses.

Total Prob = 1 * 10/11 * 8/10 * 6/9 = 16/33
  [#permalink] 08 Aug 2007, 01:28
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