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...=)
 A: Convince yourself, that there are essentially only two ways of arranging the people: Either in a big circle or in two groups of three people.
Fix one person, then:


*

*In the first case, there are $5!$ ways to arrange the five people in a circle around this person. As symmetry does not count, you get $\frac 12\cdot  5!=60$ ways in this case.

*Second case: There are $\binom{5}{2}=10$ ways to choose the people to hold hands with. The other pairings are uniquely determined by this choice.
This yields $70$ in total.
A: Think of the six people as sitting around a table, holding hands with the person to their left and right. If everyone moved one seat to the left, you'd still have the same "pattern" of hand-holding, so we want to not allow this. To make that impossible, let's put person 6 at the Northmost seat of the round table, and not let her move. 
Then all that remains is to decide where to put the other 5 people: each one can sit in any one of the 5 seats. So there are 5 places for person 1 to sit; 4 left for person 2; 3 left for person 3, and so on, for a total of $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ possible arrangements. 
But that's not QUITE the whole story: for every arrangement, we could flip the table east-west and have an equivalent arrangement: if person 6 had 3 to her left and person 1 to her right, after the flip she'd have person 3 to her right and person 1 to her left. If you call this the "same" arrangement (in the sense that person 6 is holding hands with 1 and 3 in both cases), then you have to divide by 2 to get 60 possibilities. If you call it a different arrangement (because the person to person 6's left is different in the two situations), then your answer is 120. 
A: 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 $60 + 10 = 70$ 
