Problem with concepts of circular permutation. I am having problem in understanding this concept:  

Circular permutation : The definition in my book goes like that ' Arrangements of things in a circle or a ring are called circular permutations. The fundamental difference between linear and that of circular permutation is that in the former, there are always two separate ends but in circular permutations we cannot distinguish the two ends. For this,in linear permutations ,arrangements depend on the absolute position while in the case of circular permutations,we shall be concerned with relative positions of the things. Thus, no. of circular permutations of $n$ different things taken all at a time is $$(n-1)!$$ ways taking one of the $n$ things fixed. 

This is what is written in the book. Now, it gave two questions ; 1. In how many ways can 6 boys form a ring? . The answer was,according to the formula, was $5!$ ways. 2. In how many ways can 6 men be seated at a round table? The answer, I thought,would be again $5!$ . But book gave the answer $6!$ giving reason that since $6$ men were to be arranged with respect to the table and not with themselves,hence the problem is equivalent to linear permutation.
But I didn't understand their reasoning; 6 men were forming a ring and sitting around a round table,what is the difference? And what the book wanted to tell by saying relative position ? 
Another problem is the necklace problem: In how many ways can 6 beads of different colours be arranged to form necklace? The answer,again to me, was $5!$ . But the correct answer ,the book said,was $\frac{5!}{2}$ reasoning that clockwise & anti-clockwise arrangement can't be distinguished. But I couldn't understand this brief reasoning. 
Plz help explaining these problems.   
 A: The "ring" question asks "how many ways are there to put 6 people in a ring", where two "ways" are identical if each individual has the same left and right neighbours.
The "table" question asks "how many ways are there to put 6 people at a table", where two "ways" are identical if each individual has the same left and right neighbours and the same place at the table.
It is easy to see that since the possible rotations are 6, the answer to the first question will be 6 times the answer to the second question, and indeed 6! = 6 x 5!.
Concerning the "necklace" problem, the difference from the "ring" problem is that you cannot put people upside down without noticing - while you can to so with a necklace, obtaining a pattern which is symmetric. So the number of combination is halved, because two symmetric combinations are considered as one.
A: The way I like to thing of the ring situation is as follows: Let's say four people A, B, C, D want to sit in a circle. For any one of the seating arrangements, there are four "rotational possibilities". For example, if let's say A is next to B, B next to C, C next to D and D next to A, then the "rotational possibilities" are (think of looking at the table from the top): 


*

*A at the bottom, B right, C top, D right

*A right, B top, C left, D bottom

*A top, B left, C bottom, D right

*A left, B, bottom, C right, D top.


But remember that all these are the same because of this being a circular arrangement, therefore you divide by 4 to avoid counting each four times instead of once. So in this case what we get is $4!/4 = 3!$ which is the same as your book's $(n-1)!$
As for the table/ring problem, I don't see any difference in the two cases, but I guess if you want to count these rotational possibilities separately then your book's answer is right. However I wouldn't interpret it like your book did. 
A: When you arrange 6 men in a ring, what matters is the position of one of these men respect to the other (imagine that each one takes the hand of his neighbor) , there is no absolute position with respect to which we can take into account the rotation. This is different in the case of table.
