For instance if we have a necklace with different coloured beads (it is flippable) why should we divide $(n-1)!$ by two, meaning we are not even considering the permutations of the flipped necklace (to be extra) in the first place to be cancelled out of the total permutations.
I have a small idea about it though....I think it is because we get 2 permutations by arranging objects circularly in just one way, just by flipping it...not sure though
In general when counting permutations in a circle, a rotation of any one permutation is not considered to be a different permutation; thus the position of the first bead is immaterial, and so the number of permutations is $\,(n-1)!\,$
On top of that, your necklaces are "flippable", which implies that a reflection of any one permutation is not considered to be a different permutation; since every permutation has two reflections, we have to reduce our count by half, so our final answer will be $\,\frac{(n-1)!}{2}.\,$
Note that the last step assumes that there is indeed a reflection for each permutation, which isn't true if the necklace only has one or two beads, so the formula only holds for $\,n\ge 3;\,$ for $\,n=1\,$ and $\,n=2,\,$ the correct number would just be $\,(n-1)!=1.\,$[My thanks to N.F. Taussig for pointing out the need for this addendum.]