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