Ok say there are 7 guys A B C D E F G

Now 4 have exactly 1 friend and 3 have exactly 2

So the total number of pairs of friends:

A-B
C-D
E-F+G
F-E+G
G-E+F

That makes it a total of 5 pair of friends.

Total number of ways of selecting 2 people out of 7 = 7C2 = 21

Hence probability of selecting 2 people who are NOT friends:

1-probability of selecting 2 friends = 1-5/21 = 16/21

Mals !!
How u arrived on pairs of friends...pls elaborate

See there are 7 people in the party

A B C D E F G

Now exactly 4 people have 1 friend

Lets say A B C D have exactly 1 friend

A is B's friend
C is D's friend

Now in the question its given that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend

This means if A is B's friend, then B is also A's friend and since we assumed that B has only 1 friend A will be B's only friend right.

So we have 1 pair of friendship A-B
Same logic for C and D

So that gives us 2 pairs, A-B; C-D

Now exactly 3 people have 2 friends

Lets say E F G each has exactly 2 friends. But neither one of them can be friends with A, B, C and D (because they are done with their share of friendship)

So E will have 2 friends F and G
F will have 2 friends E and G
G will have 2 friends E and F

So now we have 3 more pairs.

E = F+G
F = E+G
G = F+E

So in total 5 pairs.

Still confused
:

Mals24 in da house!!! Beautiful solution.

Logitech,

What should be the approach to solve these type of questions if the numbers are too large??? I mean here there are 7 people. What if the question asks about 20 people?
Is there any approach we shall follow?