Amardeep Sharma wrote:

Isnt the answer E 16/21

u r right, man! -)

check cicerone's solution out

http://www.gmatclub.com/phpbb/viewtopic ... ly++friendI'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

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