Necklace: Combinatorics There are $n$ distinct beads. The number of arrangements when forming a necklace is $\frac{(n-1)!}{2}$.
I am not able to understand, why we have to divide it by $2$.
They say that turning it upside down, won't make any difference. But I don't get it. 
 A: I'll try to give some explanation to you about the answer.
First note that if we have $n$ beads, then for the first bead there are $n$ option, for the next $n-1$, for the third $n-2$ and so on. So there are $n!$ permutations.
But note that the necklace is a loop, and it doesn't matter from which bead we start counting. So the combinations $ABCD$ and $BCDA$ for 4 beads are completely the same. But because there are $n$ beads, there are $n$ possibilites to start counting from. So now the total number of arrangements will be:
$$\frac{n!}{n} = (n-1)!$$
For this we could think like fixing the bead $A$ as out starting point so we need just to find permutations of the other $n-1$ beads.
The reason why it's divided by 2 is because there are 2 way to count. For example $ABCD$ and $ADCB$ are completely the same, because we could start counting in one direction and we'll get the first combinations and if we start to count in the other direction we'll get the second direction. So the total number of distinguishable arrangements is:
$$\frac{(n-1)!}{2}$$
