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# Picking up friends ....Permutation Combination

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Manager
Joined: 02 Sep 2008
Posts: 102
Picking up friends ....Permutation Combination  [#permalink]

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19 Jan 2009, 23:38
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In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

--== Message from GMAT Club Team ==--

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Manager
Joined: 05 Jul 2008
Posts: 131
GMAT 2: 740 Q51 V38
Re: Picking up friends ....Permutation Combination  [#permalink]

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20 Jan 2009, 03:17
milind1979 wrote:
In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21
SVP
Joined: 04 May 2006
Posts: 1774
Schools: CBS, Kellogg
Re: Picking up friends ....Permutation Combination  [#permalink]

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20 Jan 2009, 05:52
DavidArchuleta wrote:
In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21
milind1979 wrote:
In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

Group A: 4 people have exactly 1 friend in the room and Group B: 3 people have exactly 2 friends in the room

I pick 1 person. He could be member of Group A or B
The possibility of picking a member from Group A is: 4/7. The possibility of picking a person from 6 left that is not the first person's friend is 5/6.

The possibility of picking a member from Group B is 3/7. The possibility of picking another from 6 left that is not the first one's friend is 4/6

so the result IMO is:
4/7 * 5/6 + 3/7 * 4/6 = 16/21

[ Post details ]

Good job! DavidArchuleta
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Joined: 07 Nov 2007
Posts: 1728
Location: New York
Re: Picking up friends ....Permutation Combination  [#permalink]

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20 Jan 2009, 08:59
milind1979 wrote:
In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

A B C D --group 1 who have only one friend
E F G -- group 2 who have exactly two friends

try to find possible ways two selected are friends
AE
BE
CG
DF
FG
only 5 combinations possible

p= 1- 5/21 = 16/21

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: Picking up friends ....Permutation Combination &nbs [#permalink] 20 Jan 2009, 08:59
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