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# In a room filled with 7 people, 4 people have exactly 1

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Manager
Joined: 22 Feb 2007
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In a room filled with 7 people, 4 people have exactly 1 [#permalink]  17 Mar 2007, 17:16
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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?
5/21
3/7
4/7
5/7
16/21
Director
Joined: 13 Dec 2006
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Location: Indonesia
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number of ways of choosing 2 people out of 7 is 7c2 = 21

we know that there are 4 people who have only 1 freind, while rest three people have 2 freinds each.

this can be done only in one way, where out of these 4 people two will be attached to 1 person having two freinds while rest two will be freinds of the other two peope having 2 freinds each and are also each other's freinds. (grid will be better to explain)

so we can choose 2 people out of 4 having only one freind in 4c2 ways = 6
also the person having freindship with the people having 1 freind each will not have freindship with other 4, so we can also choose them in 4 ways.

and rest of the two people having two freinds each will not have freindship with 3 others having only one freind each. which will give us number 6.

adding them we will get number of ways of choosing 2 people not being freinds is 16.

so the probability is 16/21.

My solution is bit rustic... can somebody bring a short cut approach.

regards,

Amardeep
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Amardeep Sharma wrote:

u r right, man! -)

check cicerone's solution out http://www.gmatclub.com/phpbb/viewtopic ... ly++friend

I'm afraid to get such so complicated task in exam.

cicerone wrote:
Folks, let's keep this simple.

A,B,C,D,E,F,G are the seven friends.

Remeber if A is a friend to B, it implies B is also a friend to A.

It is given that There are exactly 4 people having 1 friend.
So we need two pairs to satisfy this
let it be
A----B
C----D

So it is clear that A ,B,C,D r the 4 pople having 1 friend only.
Now we cannot select any one of these 4 for the other combinations(bcoz in that case they will have more than 1 friend)

It is given that 3 people have exactly 2 friends.
So we need 3 pairs from remaing (E,F,G)

ie E-----F
E----G
F----G

Clearly E have only 2 friends.
SimilarlyF and G also have only 2 friends.

So totally there are 5 pairs of friends.

Now 2 people from 7 can be selected in 7C2 ie 21 ways.
Out of these 21 there r 5 pairs of friends.
Excluding them we have another 16 pairs.

So prob = 16/21

I think i am clear
[/quote]
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