Combination on a cycle? 
A bracelet is to be made by threading four identical red beads and four identical yellow beads onto a hoop. How many different bracelets can be made?

I imagine first to pick up one bead as a start, then count the number of combinations of the remaining beads, which is $$ {7!} \over {3!4!}$$
Since the beads are arranged in a cycle, I should divide this number by 2, but this turns out to be a fraction.
After checking my solution a number of times, I still cannot see what is wrong with my approach.
Could you please point out to me where I got it wrong?
 A: If you represent a red bead with $1$ and a yellow bead with $0$ then a list like $10011100$ represents a bracelet. How many of them? And given a bracelet, how many bracelets are equal to it?
It is not easy, however, to count out equal bracelets: it is probably simpler to list them all. There are only 10 different bracelets and the pattern should be clear:
00001111 
00010111
00011011
00011101
00100111
00101011
00101101
00110011
00110101
01010101
In addition, if you consider that a bracelet can be flipped over, n. 2 and 4 count as a single bracelet. The same holds for n. 6 and 9, so that in the end we are left with 8 bracelets: 
00001111 
00010111
00011011
00100111
00101011
00101101
00110011
01010101
A: Here is another way of thinking.
Denote the bracelet by $p|q|r|... $ where p is the number of beads of one colour, q is the number of beads of another colour, and r is the number of beads of the first colour and so on.
The colours alternate so there must be an even number of section
If the number of section is 2, then $4|4$ is one way.
If the number of section is 4, $2|2|2|2, \ 2|3|1|2, \ 3|3|1|1,3|1|3|1$ are the four ways of making the bracelet.
If the number of section is 6, $2|2|1|1|1|1, \ 2|1|1|2|1|1$ are two ways.
If the number of section is 8, $1|1|1|1|1|1|1|1$ is one way.
So in total there are 8 different bracelets that can be made.
