Problem Solving

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?

Solution: 2 people can be selected out of 7 in 7c2 = 21 ways

now 4 people have exactly one friend i.e. they will constitute 2 combinations e.g AB CD (mutually 4 people with 1 friend each)

& 3 with two friends will make 3 combinations i.e EF FH HE

total =5 combinations

so probability that both are friends is 5/21

so P (both are not friends ) =1-5/21 = 16/21

I DINT UNDERSTAND THE MUTUAL FRIENDSHIP THG AND THOSE 5 COMBINATIONS. PLZ ELABORATE AND EXPLAIN.

suppose the 7 people are A,B,C,D,E,F,G

here possible groups of 2 friends are

AB (A is a friend of B & B is a friend of A this is the mutual friendship i.e only one combination represents two friendships i.e between A & B , B & A)
CD

EF (here E is a friend of F & G, G of E & F & F of E & G i.e each individual has two friends but only three combinations represent this as the friendship is mutual(works from both sides) as stated above)
EG
GF

Hope this helps

Is the answer 4/7?

Total N = 7C2
n = 4C1*3C1

Therefore, P = n/N = 4/7

OA is 16/21..

how doea 4 people having 1 friend constitute 2 combinations, please explain