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In a room filled with 7 people, 4 people have exactly 1 [#permalink]
21 May 2007, 23:24

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?

1 and 2 are friends (each has exactly 1 friend)
3 and 4 are friends (each has exactly 1 friend)
5, 6 and 7 are friends (each has exactly 2 friends)

Total possible number of outcomes of picking 2 random ppl from 7 is 7!/5!2! = 21

If 1 is randomly picked, than there is only 1 outcome that another randomly picked will be his friend.
The same goes for number 2,3 and 4
For numbers 5,6 and 7 there are 2 outcomes respectivly. In total there are 10 "favourable" outcomes. Means that 11 are "unfavourable". But there is no 11/21 among answers:) So, I must have done a mistake in my last part of calculations

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