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In a room filled with 7 people, 4 people have exactly 1 [#permalink]
17 Jul 2006, 16:47

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. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

5/21
3/7
4/7
5/7
16/21

Can someone please show me a quick way to approach this question. I figured it out but it's taken me way too long..

Total selections = 21
Consider A B C D E F G
There can be 5 set of friends
consider AB CD EF FG GE
prob of choosing friends = 5/21
not choosing friends = 16/21 (E)