I do not know how to start this problem-help needed There are 6 people who are holding hands, such that each person is holding hands with exactly 2 other people. How many ways are there for them to do that?
My friend challenged me to this problem and i dont know where to start...
Thanks for any help...=)

Assuming we don't care which hand is being used to do the holding then with three people there is only one way.  You add a fourth person and he/she can go into 1 of 3 positions making 3.
Another (fifth) person and they have 4 positions to choose from so $4 \times 3 = 12$.
Another (sixth) person and they have 5 positions to choose from so $5 \times 12 = 60$.
This assumes they are all in one group If not i.e two groups of three.  Then within those two groups there is only way they can  hold each others hands but there are $6 \times 5 \times 4 =120$ ways you can pick these groups. We can divide this by two because if Ann, Bob and Carol are holding hands we don't care which table they are at. Because of the way we picked these groups ABC and ACB are counted as separate groups and we don't want to count these twice so we can divide this by 6 to get 10.
So the answer is $60 + 10 = 70$ 
